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For this Question: You are given a big number in form of a string A of characters from 0 to 9.
Check whether the given number is divisible by 30 .The first argument is the string A.
<b>Constraints</b>
1 ≤ |A| ≤ 10<sup>5</sup>Return "Yes" if it is divisible by 30 and "No" otherwise. Sample Input :
3033330
Sample Output:
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
String a= br.readLine();
int sum=0;
int n=a.length();
int s=0;
for(int i=0; i<n; i++){
sum=sum+ (a.charAt(i) - '0');
if(i == n-1){
s= (a.charAt(i) - '0');
}
}
if(sum%3==0 && s==0){
System.out.print("Yes");
}else{
System.out.print("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a big number in form of a string A of characters from 0 to 9.
Check whether the given number is divisible by 30 .The first argument is the string A.
<b>Constraints</b>
1 ≤ |A| ≤ 10<sup>5</sup>Return "Yes" if it is divisible by 30 and "No" otherwise. Sample Input :
3033330
Sample Output:
Yes, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-10 11:06:49
**/
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int solve(string a) {
int n = a.size();
int sum = 0;
if (a[n - 1] != '0') {
return 0;
}
for (auto &it : a) {
sum += (it - '0');
}
debug(sum % 3);
if (sum % 3 == 0) {
return 1;
}
return 0;
}
int32_t main() {
ios_base::sync_with_stdio(NULL);
cin.tie(0);
string str;
cin >> str;
int res = solve(str);
if (res) {
cout << "Yes\n";
} else {
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a big number in form of a string A of characters from 0 to 9.
Check whether the given number is divisible by 30 .The first argument is the string A.
<b>Constraints</b>
1 ≤ |A| ≤ 10<sup>5</sup>Return "Yes" if it is divisible by 30 and "No" otherwise. Sample Input :
3033330
Sample Output:
Yes, I have written this Solution Code: A=int(input())
if A%30==0:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N cities on X-axis and you want to visit all of them. For traveling you can walk or teleport.
For 1 unit walk the cost is A unit. The price is B units for teleportation from any point to any other.
Find the minimum possible total cost when you visit all the cities. Initially, you are at city 1.The first line of input contains three space-separate integers N, A, B
The second input line includes N distinct space-separated integers X1, X2,.....Xn which are the coordinates of the city on the X-axis.
<b>Constraints:-</b>
1 <= N <= 10<sup>5</sup>
1 <= A,B <= 10^9
1 <= Xi <= 10^9
Xi < X(i+1)Print the minimum possible total cost when you visit all the cities.Sample Input :
4 2 5
1 2 5 7
Sample Output :
11
<b>Explanation:</b>
From town 1, walk a distance of 1 to town 2, then teleport to town 3, then walk a distance of 2 to town 4. The total increase of your fatigue level, in this case, is 2×1 + 5 + 2×2 = 11, which is the minimum possible value., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define YES(j) cout << (j ? "YES" : "NO") << endl;
#define Yes(j) std::cout << (j ? "Yes" : "No") << endl;
#define yes(j) std::cout << (j ? "yes" : "no") << endl;
typedef long long ll;
int main(void) {
#ifdef ANIKET_GOYAL
freopen("inputf.in", "r", stdin);
freopen("outputf.in", "w", stdout);
#endif
long long n, a, b;
cin >> n >> a >> b;
assert(1 <= n <= 100000);
assert(1 <= a <= 1000000000);
assert(1 <= b <= 1000000000);
long long ans = 0;
int pos = 0;
int prev;
REP(i, n) {
int x;
cin >> x;
assert(1 <= x <= 1000000000);
if (i) {
if (x < prev) {
cout << prev << " " << x << "\n";
return 0;
}
}
if (i == 0) {
pos = x;
} else {
ans += min(a * (x - pos), b);
pos = x;
}
prev = x;
}
cout << ans << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N cities on X-axis and you want to visit all of them. For traveling you can walk or teleport.
For 1 unit walk the cost is A unit. The price is B units for teleportation from any point to any other.
Find the minimum possible total cost when you visit all the cities. Initially, you are at city 1.The first line of input contains three space-separate integers N, A, B
The second input line includes N distinct space-separated integers X1, X2,.....Xn which are the coordinates of the city on the X-axis.
<b>Constraints:-</b>
1 <= N <= 10<sup>5</sup>
1 <= A,B <= 10^9
1 <= Xi <= 10^9
Xi < X(i+1)Print the minimum possible total cost when you visit all the cities.Sample Input :
4 2 5
1 2 5 7
Sample Output :
11
<b>Explanation:</b>
From town 1, walk a distance of 1 to town 2, then teleport to town 3, then walk a distance of 2 to town 4. The total increase of your fatigue level, in this case, is 2×1 + 5 + 2×2 = 11, which is the minimum possible value., I have written this Solution Code: import java.util.*;
public class Main
{
public static void main (String [] args)
{
Scanner scan = new Scanner (System.in);
int n = scan.nextInt();
int a = scan.nextInt();
int b = scan.nextInt();
int [] cities = new int [n];
// int resArr [] = new int [n];
for (int i=0; i<n; i++)
{
cities[i] = scan.nextInt();
}
//Arrays.sort(cities);
long sum =0;
// if(cities[0] != 1)
// {
// int diffInCities = cities[0]-1;
// int walkCost = a * diffInCities;
// int teleportCost = b;
// sum = sum + (walkCost>teleportCost?teleportCost:walkCost);
// }
for(int i=1; i<n; i++)
{
int diffInCities = cities[i]-cities[i-1];
int walkCost = a * diffInCities;
int teleportCost = b;
sum = sum + (walkCost>teleportCost?teleportCost:walkCost);
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an undirected unweighted connected graph consisting of N vertices (numbered 1 to N) and N edges. Let d(u, v) denote the shortest distance between nodes u and v in the graph. You need to find the sum of d(u, v)<sup>K</sup> over all pairs (u, v) such that 1 ≤ u < v ≤ N. Since this sum can be large, you need to print it modulo 998244353.
Note: There can be self-edges in the graph.The first line contains two space-separated integers N and K.
Then N lines follow, each containing two space-separated integers u and v, denoting an edge between vertices u and v.
<b> Constraints: </b>
1 ≤ N ≤ 2×10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ u, v ≤ NPrint a single integer, the summation of d(u, v)<sup>K</sup> modulo 998244353.Sample Input 1:
3 1
1 2
2 3
3 1
Sample Output 1:
3
Sample Explanation 1:
Answer = d(1,2) + d(1,3) + d(2,3) = 1 + 1 + 1 = 3
Sample Input 2:
4 2
1 2
1 3
1 4
1 1
Sample Output 2:
15
Sample Explanation 2:
Answer = d(1,2)<sup>2</sup> + d(1,3)<sup>2</sup> + d(1,4)<sup>2</sup> + d(2,3)<sup>2</sup> + d(2,4)<sup>2</sup> + d(3,4)<sup>2</sup> = 1<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup> + 2<sup>2</sup> + 2<sup>2</sup> + 2<sup>2</sup> = 15, I have written this Solution Code: #include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
typedef vector<int> Poly;
const int MAXN=1e5+5,mod=998244353,G=3,invG=(mod+1)/3;
int n,K,ed[MAXN][2],m,pre[MAXN],lp[MAXN],tl[MAXN],m2;
Poly ans;
int fnd(int x){
if(x!=pre[x]) pre[x]=fnd(pre[x]);
return pre[x];
}
int cnte,h[MAXN],to[MAXN<<1],nx[MAXN<<1];
inline void adde(int u,int v){
cnte++;
nx[cnte]=h[u];
to[cnte]=v;
h[u]=cnte;
}
ll Fstpw(ll a,int b){
ll res=1;
while(b){
if(b&1) res=res*a%mod;
b>>=1;
a=a*a%mod;
}
return res;
}
void ntt(int *a,int n,int tp){
int bit=0;
while(1<<bit<n) bit++;
static int rev[MAXN<<2];
for(int i=1; i<n; i++){
rev[i]=rev[i>>1]>>1|((i&1)<<bit-1);
if(i<rev[i]) swap(a[i],a[rev[i]]);
}
for(int mid=1; mid<n; mid<<=1){
ll w=1,w1=Fstpw(tp==1?G:invG,(mod-1)/mid/2);
for(int j=0; j<mid; j++,w=w*w1%mod)
for(int i=0; i<n; i+=mid*2){
int x=a[i+j],y=w*a[i+j+mid]%mod;
a[i+j]=(x+y)%mod;
a[i+j+mid]=(x-y+mod)%mod;
}
}
if(tp==-1){
ll t=Fstpw(n,mod-2);
for(int i=0; i<n; i++)
a[i]=a[i]*t%mod;
}
return ;
}
Poly operator +(Poly a,Poly b){
if(a.size()<b.size()) swap(a,b);
for(int i=0; i<b.size(); i++)
a[i]=(a[i]+b[i])%mod;
return a;
}
Poly operator -(Poly a,Poly b){
if(a.size()<b.size()) a.resize(b.size());
for(int i=0; i<b.size(); i++)
a[i]=(a[i]-b[i]+mod)%mod;
return a;
}
//long clk;
Poly operator *(Poly a,Poly b){
//clk-=clock();
if(a.empty()||b.empty()) return Poly{};
int n=a.size()-1,m=b.size()-1,siz=1;
while(siz<=n+m) siz<<=1;
static int f[MAXN<<2],g[MAXN<<2];
for(int i=0; i<siz; i++){
f[i]=i<=n?a[i]:0;
g[i]=i<=m?b[i]:0;
}
ntt(f,siz,1);
ntt(g,siz,1);
for(int i=0; i<siz; i++)
f[i]=1ll*f[i]*g[i]%mod;
ntt(f,siz,-1);
//clk+=clock();
return Poly(f,f+n+m+1);
}
inline void ad(Poly &c,Poly &a,Poly &b,int s){
Poly t(a*b);
if(c.size()<t.size()+s) c.resize(t.size()+s);
for(int i=0; i<t.size(); i++)
c[i+s]=(c[i+s]+t[i])%mod;
return ;
}
inline void ad(Poly &c,Poly &a,int s){
if(c.size()<a.size()+s) c.resize(a.size()+s);
for(int i=0; i<a.size(); i++)
c[i+s]=(c[i+s]+a[i])%mod;
return ;
}
int fa[MAXN],dep[MAXN],siz[MAXN];
Poly a[MAXN];
void Dfs1(int u){
for(int i=h[u]; i; i=nx[i]){
int v=to[i];
if(v==fa[u]) continue;
fa[v]=u;
dep[v]=dep[u]+1;
Dfs1(v);
}
return ;
}
int cen;
bool vis[MAXN];
int getrt(int u,int s){
siz[u]=1;
int mx=0;
for(int i=h[u]; i; i=nx[i]){
int v=to[i];
if(vis[v]||v==fa[u]) continue;
fa[v]=u;
getrt(v,s);
mx=max(mx,siz[v]);
siz[u]+=siz[v];
}
mx=max(mx,s-siz[u]);
if(mx<=s/2) cen=u;
return cen;
}
Poly cnt;
void Add(int co){
cnt=cnt*cnt;
if(cnt.size()>n) cnt.resize(n);
for(int i=1; i<cnt.size(); i++)
ans[i]=(ans[i]+1ll*cnt[i]*co+mod)%mod;
cnt.clear();
return ;
}
void Dfs2(int u){
siz[u]=1;
if(dep[u]<cnt.size()) cnt[dep[u]]++;
else cnt.push_back(1);
for(int i=h[u]; i; i=nx[i]){
int v=to[i];
if(vis[v]||v==fa[u]) continue;
fa[v]=u;
dep[v]=dep[u]+1;
Dfs2(v);
siz[u]+=siz[v];
}
return ;
}
void Divide(int u){
vis[u]=1;
fa[u]=0;
dep[u]=0;
Dfs2(u);
Add(1);
for(int i=h[u]; i; i=nx[i]){
int v=to[i];
if(vis[v]) continue;
cnt.push_back(0);
Dfs2(v);
Add(-1);
v=getrt(v,siz[v]);
Divide(v);
}
return ;
}
Poly f[MAXN<<2],g[MAXN<<2];
#define lc k<<1
#define rc k<<1|1
#define ls lc,l,mid
#define rs rc,mid+1,r
#define Clear() f[lc].clear(),f[rc].clear(),g[lc].clear(),g[rc].clear()
void Dfs3(int k,int l,int r){
//for(int i=l; i<=r; i++) for(int j=i+1; j<=r; j++) ans=ans+Shift(a[i]*a[j],j-i); return ;
if(l==r){
f[k]=g[k]=a[l];
return ;
}
int mid=l+r>>1;
Dfs3(ls);
Dfs3(rs);
ad(ans,g[lc],f[rc],1);
swap(f[k],f[lc]);
ad(f[k],f[rc],mid-l+1);
swap(g[k],g[rc]);
ad(g[k],g[lc],r-mid);
Clear();
return ;
}
void Dfs4(int k,int l,int r){
//for(int i=1; i<=m2; i++) for(int j=i+1; j<=m2; j++) ans=ans+Shift(a[i+m2]*a[j],i+m2-j); return ;
if(l==r){
f[k]=a[l+m2];
g[k]=a[l];
return ;
}
int mid=l+r>>1;
Dfs4(ls);
Dfs4(rs);
ad(ans,f[lc],g[rc],l+m2-r);
swap(f[k],f[lc]);
ad(f[k],f[rc],mid-l+1);
swap(g[k],g[rc]);
ad(g[k],g[lc],r-mid);
Clear();
return ;
}
void Dfs5(int k,int l,int r){
//for(int i=1; i<=m2; i++) for(int j=i+1; j<=m2; j++) ans=ans+Shift(a[i]*a[j+m2],i+m-j-m2); return ;
if(l==r){
f[k]=a[l];
g[k]=a[l+m2];
return ;
}
int mid=l+r>>1;
Dfs5(ls);
Dfs5(rs);
ad(ans,f[lc],g[rc],l+m-r-m2);
swap(f[k],f[lc]);
ad(f[k],f[rc],mid-l+1);
swap(g[k],g[rc]);
ad(g[k],g[lc],r-mid);
Clear();
return ;
}
int main(){
//freopen("sub16.in","r",stdin);
//freopen("b.out","w",stdout);
scanf("%d%d",&n,&K);
ans.resize(n+1);
for(int i=1; i<=n; i++)
pre[i]=i;
for(int i=1; i<=n; i++){
int &u=ed[i][0],&v=ed[i][1];
scanf("%d%d",&u,&v);
int x=fnd(u),y=fnd(v);
if(x==y) ed[0][0]=u,ed[0][1]=v;
else pre[x]=y,adde(u,v),adde(v,u);
}
Dfs1(ed[0][0]);
int p=ed[0][1];
while(p!=ed[0][0]){
lp[++m]=p;
p=fa[p];
}
lp[++m]=ed[0][0];
for(int i=1; i<=m; i++)
tl[lp[i]]=i,vis[lp[i]]=1;
for(int i=1; i<=n; i++){
int u=i;
while(!tl[u]) u=fa[u];
int t=tl[u];
u=i;
while(!tl[u]){
tl[u]=t;
u=fa[u];
}
int d=dep[i]-dep[lp[t]];
if(d>=a[t].size()) a[t].resize(d+1);
a[t][d]++;
}
//printf("%.2f\n",(double)(clock())/CLOCKS_PER_SEC);
for(int i=1; i<=m; i++)
Divide(lp[i]);
//printf("%.2f\n",(double)(clock())/CLOCKS_PER_SEC);
for(int i=1; i<=n; i++)
ans[i]=ans[i]*(mod+1ll)/2%mod;
if(m>1){
//for(int i=1; i<=m; i++) for(int j=i+1; j<=m; j++) ans=ans+Shift(a[i]*a[j],min(j-i,i+m-j));
m2=m/2;
for(int i=1; i<=m2; i++)
ad(ans,a[i],a[i+m2],m2);
//printf("%.2f\n",(double)(clock())/CLOCKS_PER_SEC);
Dfs3(1,1,m2);
//printf("3 %.2f\n",(double)(clock())/CLOCKS_PER_SEC);
Dfs3(1,m2+1,m2*2);
//printf("3 %.2f\n",(double)(clock())/CLOCKS_PER_SEC);
Dfs4(1,1,m2);
//printf("4 %.2f\n",(double)(clock())/CLOCKS_PER_SEC);
Dfs5(1,1,m2);
//printf("5 %.2f\n",(double)(clock())/CLOCKS_PER_SEC);
if(m&1){
Poly t(n);
for(int i=1; i<=m2; i++){
for(int j=0; j<a[i].size(); j++)
t[i+j]=(t[i+j]+a[i][j])%mod;
for(int j=0; j<a[i+m2].size(); j++)
t[m-i-m2+j]=(t[m-i-m2+j]+a[i+m2][j])%mod;
}
ad(ans,t,a[m],0);
}
//printf("odd %.2f\n",(double)(clock())/CLOCKS_PER_SEC);
}
int as=0;
for(int i=1; i<n; i++)
as=(as+Fstpw(i,K)*ans[i])%mod;
as=as%mod;
printf("%d\n",as);
//printf("%.2f\n",(double)(clock())/CLOCKS_PER_SEC);
//printf("ntt %.2f\n",(double)(clk)/CLOCKS_PER_SEC);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and X, where N is the number of total patients and X is the time duration (in minutes) after which a new patient arrives. Also, doctor will give only 10 minutes to each patient. The task is to calculate the time (in minutes) the last patient needs to wait.The first line of input contains the number of test cases T.
The next T subsequent lines denote the total number of patients N and time interval X (in minutes) in which the next patients are visiting.
Constraints:
1 <= T <= 100
0 <= N <= 100
0 <= X <= 30Output the waiting time of last patient.Input:
5
4 5
5 3
6 5
7 6
8 2
Output:
15
28
25
24
56, I have written this Solution Code: for i in range(int(input())):
n, x = map(int, input().split())
if x >= 10:
print(0)
else:
print((10-x)*(n-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and X, where N is the number of total patients and X is the time duration (in minutes) after which a new patient arrives. Also, doctor will give only 10 minutes to each patient. The task is to calculate the time (in minutes) the last patient needs to wait.The first line of input contains the number of test cases T.
The next T subsequent lines denote the total number of patients N and time interval X (in minutes) in which the next patients are visiting.
Constraints:
1 <= T <= 100
0 <= N <= 100
0 <= X <= 30Output the waiting time of last patient.Input:
5
4 5
5 3
6 5
7 6
8 2
Output:
15
28
25
24
56, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n, x;
cin >> n >> x;
if(x >= 10)
cout << 0 << endl;
else
cout << (10-x)*(n-1) << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and X, where N is the number of total patients and X is the time duration (in minutes) after which a new patient arrives. Also, doctor will give only 10 minutes to each patient. The task is to calculate the time (in minutes) the last patient needs to wait.The first line of input contains the number of test cases T.
The next T subsequent lines denote the total number of patients N and time interval X (in minutes) in which the next patients are visiting.
Constraints:
1 <= T <= 100
0 <= N <= 100
0 <= X <= 30Output the waiting time of last patient.Input:
5
4 5
5 3
6 5
7 6
8 2
Output:
15
28
25
24
56, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine());
while (T -->0){
String s[] = br.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int p = Integer.parseInt(s[1]);
if (p<10)
System.out.println(Math.abs(n-1)*(10-p));
else System.out.println(0);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and a matrix of size (N x N). Check whether given matrix is identity or not.
<b> Note </b>
An identity matrix is a square matrix in which all the elements of principal diagonal are one, and all other elements are zeros.First line contain a single integer N
Next N line contain N space- separated integer i. e. elements of matrix.If the given matrix is the identity matrix. print "Yes" otherwise print "NO"
Constraints:
1<=N<=100Sample Input 1:
3
1 0 0
0 1 0
0 0 1
Sample Output 1:
Yes
Explanation:
Given matrix is an identity matrix because all main diagonal elements are 1 and rest of elements are 0., I have written this Solution Code: import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[] ) throws Exception {
Scanner s = new Scanner(System.in);
int N = s.nextInt();
int [][]arr=new int[N][N];
for(int i=0;i<N;i++){
for(int j=0;j<N;j++)
arr[i][j] = s.nextInt();
}
boolean flag=false;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++)
{
if(i==j && arr[i][j]!=1)flag=true;
else if(i!=j && arr[i][j]!=0)flag=true;
}
}
if(flag==true){
System.out.println("No");
}
else {
System.out.println("Yes");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a doubly linked list consisting of N nodes and two integers <b>P</b> and <b>K</b>. Your task is to add an element K at the Pth position from the start of the linked list<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>insertnew()</b>. The description of parameters are mentioned below:
<b>head</b>: head node of the double linked list
<b>K</b>: the element which you have to insert
<b>P</b>: the position at which you have insert
Constraints:
1 <= P <=N <= 1000
1 <=K, Node.data<= 1000
In the sample Input N, P and K are in the order as mentioned below:
<b>N P K</b>Return the head of the modified linked list.Sample Input:-
5 3 2
1 3 2 4 5
Sample Output:-
1 3 2 2 4 5, I have written this Solution Code: public static Node insertnew(Node head, int k,int pos) {
int cnt=1;
if(pos==1){Node temp=new Node(k);
temp.next=head;
head.prev=temp;
return temp;}
Node temp=head;
while(cnt!=pos-1){
temp=temp.next;
cnt++;
}
Node x= new Node(k);
x.next=temp.next;
temp.next.prev=x;
temp.next=x;
x.prev=temp;
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: n=int(input())
a=map(int,input().split())
b=[]
mx=-200000
cnt=0
for i in a:
if i>mx:
cnt+=1
mx=i
print(cnt), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: function numberOfRoofs(arr)
{
let count=1;
let max = arr[0];
for(let i=1;i<arrSize;i++)
{
if(arr[i] > max)
{
count++;
max = arr[i];
}
}
return count;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++){
a[i]=s.nextInt();
}
int count=1;
int max = a[0];
for(int i=1;i<n;i++)
{
if(a[i] > max)
{
count++;
max = a[i];
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have two integers, X and Y. Initially, they are equal to A and B, respectively. In one move, you can do exactly one of the following operations:
1. Set X = X+1
2. Set Y = Y+1.
3. Let d be any divisor of X, then set X = d
4. Let d be any divisor of Y, then set Y = d.
Find the minimum number of moves so that X = Y.The input consists of two space separated integers A and B.
Constraints:
1 ≤ A, B ≤ 10<sup>9</sup>Print a single integer, the minimum number of moves so that X = Y.Sample 1:
1 8
Output 1:
1
Explanation 1:
We do only one operation, we let Y = 1 by the fourth operation. Then X = Y = 1.
Sample 2:
5 5
Output 2:
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
String s[]=bu.readLine().split(" ");
int a=Integer.parseInt(s[0]),b=Integer.parseInt(s[1]);
int ans=2;
if(a==b) ans=0;
else if(a%b==0 || b%a==0 || a+1==b || b+1==a) ans=1;
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have two integers, X and Y. Initially, they are equal to A and B, respectively. In one move, you can do exactly one of the following operations:
1. Set X = X+1
2. Set Y = Y+1.
3. Let d be any divisor of X, then set X = d
4. Let d be any divisor of Y, then set Y = d.
Find the minimum number of moves so that X = Y.The input consists of two space separated integers A and B.
Constraints:
1 ≤ A, B ≤ 10<sup>9</sup>Print a single integer, the minimum number of moves so that X = Y.Sample 1:
1 8
Output 1:
1
Explanation 1:
We do only one operation, we let Y = 1 by the fourth operation. Then X = Y = 1.
Sample 2:
5 5
Output 2:
0, I have written this Solution Code: l=list(map(int, input().split()))
A=l[0]
B=l[1]
if A==B:
print(0)
elif A%B==0 or B%A==0 or abs(A-B)==1:
print(1)
else:
print(2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have two integers, X and Y. Initially, they are equal to A and B, respectively. In one move, you can do exactly one of the following operations:
1. Set X = X+1
2. Set Y = Y+1.
3. Let d be any divisor of X, then set X = d
4. Let d be any divisor of Y, then set Y = d.
Find the minimum number of moves so that X = Y.The input consists of two space separated integers A and B.
Constraints:
1 ≤ A, B ≤ 10<sup>9</sup>Print a single integer, the minimum number of moves so that X = Y.Sample 1:
1 8
Output 1:
1
Explanation 1:
We do only one operation, we let Y = 1 by the fourth operation. Then X = Y = 1.
Sample 2:
5 5
Output 2:
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int a, b;
cin >> a >> b;
if (a == b) cout << 0;
else if (abs(a - b) == 1) cout << 1;
else if (a % b == 0 || b % a == 0) cout << 1;
else cout << 2;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a sequence of numbers of size N. You have to find if there is a way to insert + or - operator in between the numbers so that the result equals K.The first line of input contains two integers N and K. The next line of input contains N space- separated integers depicting the values of the sequence.
Constraints:-
1 <= N <= 20
-10^15 <= K <= 10^15
0 <= Numbers <=10^13Print YES if possible else print NO.Sample Input:-
4 4
1 2 3 4
Sample Output:-
YES
Sample Input:-
4 1
1 2 3 4
Sample Output:-
NO, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main {
public static boolean isArrangementPossible(long arr[],int n,long sum){
if(n==1){
if(arr[0]==sum)
return true;
else
return false;
}
return(isArrangementPossible(arr,n-1,sum-arr[n-1]) || isArrangementPossible(arr,n-1,sum+arr[n-1]));
}
public static void main (String[] args) throws IOException {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
String str1[]=br.readLine().trim().split(" ");
int n=Integer.parseInt(str1[0]);
long sum=Long.parseLong(str1[1]);
String str[]=br.readLine().trim().split(" ");
long arr[]=new long[n];
for(int i=0;i<n;i++){
arr[i]=Long.parseLong(str[i]);
}
if(isArrangementPossible(arr,n,sum)){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a sequence of numbers of size N. You have to find if there is a way to insert + or - operator in between the numbers so that the result equals K.The first line of input contains two integers N and K. The next line of input contains N space- separated integers depicting the values of the sequence.
Constraints:-
1 <= N <= 20
-10^15 <= K <= 10^15
0 <= Numbers <=10^13Print YES if possible else print NO.Sample Input:-
4 4
1 2 3 4
Sample Output:-
YES
Sample Input:-
4 1
1 2 3 4
Sample Output:-
NO, I have written this Solution Code: def checkIfGivenTargetIsPossible(nums,currSum,i,targetSum):
if i == len(nums):
if currSum == targetSum:
return 1
return 0
if(checkIfGivenTargetIsPossible(nums,currSum + nums[i],i+1,targetSum)):
return 1
return checkIfGivenTargetIsPossible(nums,currSum - nums[i], i+1,targetSum)
n,k = map(int,input().split())
nums = list(map(int,input().split()))
if(checkIfGivenTargetIsPossible(nums,0,0,k)):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a sequence of numbers of size N. You have to find if there is a way to insert + or - operator in between the numbers so that the result equals K.The first line of input contains two integers N and K. The next line of input contains N space- separated integers depicting the values of the sequence.
Constraints:-
1 <= N <= 20
-10^15 <= K <= 10^15
0 <= Numbers <=10^13Print YES if possible else print NO.Sample Input:-
4 4
1 2 3 4
Sample Output:-
YES
Sample Input:-
4 1
1 2 3 4
Sample Output:-
NO, I have written this Solution Code: #include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#define int long long
int k;
using namespace std;
int solve(int n, int a[], int i, int curr ){
if(i==n){
if(curr==k){return 1;}
return 0;
}
if(solve(n,a,i+1,curr+a[i])==1){return 1;}
return solve(n,a,i+1,curr-a[i]);
}
signed main() {
int n;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
if(solve(n,a,1,a[0])){
cout<<"YES";}
else{
cout<<"NO";}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Going through your old stuff you found a weird function F defined as :
<b>F(N) = summation of ((N-i+1)*log(i)) for i from 1 to N</b>.
Find the maximum value of b such that <b>F(N) = c*log(a) + b</b> where a, b and c are any arbitrary non negative integers satisfying the equality.
(Note that the base of log is 10 everywhere).
Since the answer can be huge, find answer % 1000000007.First line of the input contains T denoting number of test cases.
Next T lines contains a single integer N for that test case.
1 <= T <= 100
1 <= N <= 1000000000000000000 (10^18)Print the (maximum value of b) modulo 1000000007 such that F(N) = c*log(a) + b where a, b and c are non negative integers, for each test case in a single line.Sample input
4
1
2
5
12
Sample output
0
0
1
11
Explanation :
F(1) = 1*log(1) = 0
F(2) = 2*log(1) + 1*log(2) = log(2)
and so on., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
long long powerm(long long x, unsigned long long y, long long p)
{
long long res = 1;
x = x % p;
while (y > 0)
{
if (y & 1)
res = (res*x) % p;
y = y>>1;
x = (x*x) % p;
}
return res;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int t;
cin>>t;
int mo=1000000007;
while(t)
{
--t;
int n;
cin>>n;
int ans=0;
int v=5;
int ti=powerm(2,mo-2,mo);
while(v<=n)
{
int x=n/v;
ans+=(((n)%v+1)%mo)*(x%mo)%mo;
ans%=mo;
--x;
ans+=(v%mo)*(((((x%mo)*((x+1)%mo))%mo)*ti%mo))%mo;
ans%=mo;
if(v*((long double)(5))>n)
break;
v*=5;
}
cout<<ans<<"\n";
}
#ifdef ANIKET_GOYAL
cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Going through your old stuff you found a weird function F defined as :
<b>F(N) = summation of ((N-i+1)*log(i)) for i from 1 to N</b>.
Find the maximum value of b such that <b>F(N) = c*log(a) + b</b> where a, b and c are any arbitrary non negative integers satisfying the equality.
(Note that the base of log is 10 everywhere).
Since the answer can be huge, find answer % 1000000007.First line of the input contains T denoting number of test cases.
Next T lines contains a single integer N for that test case.
1 <= T <= 100
1 <= N <= 1000000000000000000 (10^18)Print the (maximum value of b) modulo 1000000007 such that F(N) = c*log(a) + b where a, b and c are non negative integers, for each test case in a single line.Sample input
4
1
2
5
12
Sample output
0
0
1
11
Explanation :
F(1) = 1*log(1) = 0
F(2) = 2*log(1) + 1*log(2) = log(2)
and so on., I have written this Solution Code: t=int(input())
for i in range(t):
number = int(input())
x = 5
ans = 0
mod = 10**9+7
while x<=number:
y = number//x
ans = (ans+y*number - x*((y*(y+1))//2) + y)%mod
x = x*5
print(int(ans)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alexa's house has only one socket.
Alexa wants to extend it with some number of power strips, each with A sockets, into B or more empty sockets. One power strip with A sockets can extend one empty socket into A empty sockets.
Find the minimum number of power strips required.The input consists of two space separated integers.
A B
<b>Constraints</b>
All values in input are integers.
2≤A≤20
1≤B≤20Print the minimum number of power strips required.<b>Sample Input 1</b>
4 10
<b>Sample Output 1</b>
3
<b>Sample Input 2</b>
8 9
<b>Sample Output 2</b>
2
<b>Sample Input 3</b>
8 8
<b>Sample Output 3</b>
1, I have written this Solution Code: #include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;
const int MOD=1000000007;
int main(){
int a,b;cin>>a>>b;b--;
a--;
cout<<(b+a-1)/a<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
public class Main {
static final int MOD = 1000000007;
public static void main(String args[]) throws IOException {
BufferedReader br
= new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(str[0]);
int b = Integer.parseInt(str[1]);
int c = Integer.parseInt(str[2]);
int d = Integer.parseInt(str[3]);
int e = Integer.parseInt(str[4]);
System.out.println(grades(a, b, c, d, e));
}
static char grades(int a, int b, int c, int d, int e)
{
int sum = a+b+c+d+e;
int per = sum/5;
if(per >= 80)
return 'A';
else if(per >= 60)
return 'B';
else if(per >= 40)
return 'C';
else
return 'D';
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: li = list(map(int,input().strip().split()))
avg=0
for i in li:
avg+=i
avg=avg/5
if(avg>=80):
print("A")
elif(avg>=60 and avg<80):
print("B")
elif(avg>=40 and avg<60):
print("C")
else:
print("D"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array consisting of N integers, your task is to reverse every subarray of K group elements.
See example for better understanding.First line of input contains two space separated integers N and K, next line contains N space separated integers containing values of array.
Constraints:-
1 < = K < = N < =100000
1 < = Arr[i] < = 100000Print the modified array.Sample Input:-
6 2
1 2 3 4 5 6
Sample Output:-
2 1 4 3 6 5
Sample Input:-
5 3
1 2 3 4 5
Sample Output:-
3 2 1 4 5
Explanation:
Step 1: 3 2 1 4 5
No more steps can be done, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int arr[] = new int [n];
int k = Integer.parseInt(str[1]);
str = read.readLine().trim().split(" ");
for(int i=0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
}
int i = 0;
while(i+k <= n){
for(int j=0;j<k/2;j++){
int temp = arr[i + j];
arr[i + j] = arr[i + k - j-1];
arr[i + k - j-1] = temp;
}
i+=k;
}
for(i=0;i<n;i++){
System.out.print(arr[i] + " ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array consisting of N integers, your task is to reverse every subarray of K group elements.
See example for better understanding.First line of input contains two space separated integers N and K, next line contains N space separated integers containing values of array.
Constraints:-
1 < = K < = N < =100000
1 < = Arr[i] < = 100000Print the modified array.Sample Input:-
6 2
1 2 3 4 5 6
Sample Output:-
2 1 4 3 6 5
Sample Input:-
5 3
1 2 3 4 5
Sample Output:-
3 2 1 4 5
Explanation:
Step 1: 3 2 1 4 5
No more steps can be done, I have written this Solution Code: n,k = map(int,input().split())
ls = list(map(int,input().split()))
for i in range(0,n,k):
if (i+k)<=n: print(*ls[i:i+k][::-1],end=' ')
else: print(*ls[i:]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array consisting of N integers, your task is to reverse every subarray of K group elements.
See example for better understanding.First line of input contains two space separated integers N and K, next line contains N space separated integers containing values of array.
Constraints:-
1 < = K < = N < =100000
1 < = Arr[i] < = 100000Print the modified array.Sample Input:-
6 2
1 2 3 4 5 6
Sample Output:-
2 1 4 3 6 5
Sample Input:-
5 3
1 2 3 4 5
Sample Output:-
3 2 1 4 5
Explanation:
Step 1: 3 2 1 4 5
No more steps can be done, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int x=n%k;
for(int i=0;i<n-x;i++){
cout<<a[(i/k)*k+(k-i%k-1)]<<" ";
}
for(int i=n-x;i<n;i++){
cout<<a[i]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: a, b = list(map(int, input().split(" ")))
print(str(a <= 10 and b >= 10).lower(), end=' ')
print(str(a % 2 == 0 or b % 2 == 0).lower(), end=' ')
print(str(not a == b).lower()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class Main {
static boolean Logical_AND(int a, int b){
if(a<=10 && b>=10){
return true;}
return false;}
static boolean Logical_OR(int a, int b){
if(a%2==0 || b%2==0){
return true;}
return false;}
static boolean Logical_NOT(int a, int b){
if(a!=b){
return true;}
return false;}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a=in.nextInt();
int b=in.nextInt();
System.out.print(Logical_AND(a, b)+" ");
System.out.print(Logical_OR(a,b)+" ");
System.out.print(Logical_NOT(a,b)+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: import java.util.InputMismatchException;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
int MAX = (int) 1e5, MOD = (int)1e9+7;
void solve(int TC) {
long n = nl();
long k = nl();
int p = 0;
while(n>0 && n%2==0) {
n/=2;
++p;
}
if(p>=k) {pn(0);return;}
k -= p;
long ans = (k+3L)/4L;
pn(ans);
}
boolean TestCases = false;
public static void main(String[] args) throws Exception { new Main().run(); }
long pow(long a, long b) {
if(b==0 || a==1) return 1;
long o = 1;
for(long p = b; p > 0; p>>=1) {
if((p&1)==1) o = (o*a) % MOD;
a = (a*a) % MOD;
} return o;
}
long inv(long x) {
long o = 1;
for(long p = MOD-2; p > 0; p>>=1) {
if((p&1)==1)o = (o*x)%MOD;
x = (x*x)%MOD;
} return o;
}
long gcd(long a, long b) { return (b==0) ? a : gcd(b,a%b); }
int gcd(int a, int b) { return (b==0) ? a : gcd(b,a%b); }
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
int T = TestCases ? ni() : 1;
for(int t=1;t<=T;t++) solve(t);
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
void p(Object o) { out.print(o); }
void pn(Object o) { out.println(o); }
void pni(Object o) { out.println(o);out.flush(); }
double PI = 3.141592653589793238462643383279502884197169399;
int ni() {
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
long nl() {
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
double nd() { return Double.parseDouble(ns()); }
char nc() { return (char)skip(); }
int BUF_SIZE = 1024 * 8;
byte[] inbuf = new byte[BUF_SIZE];
int lenbuf = 0, ptrbuf = 0;
int readByte() {
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
} return inbuf[ptrbuf++];
}
boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))) {
sb.appendCodePoint(b); b = readByte();
} return sb.toString();
}
char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
} return n == p ? buf : Arrays.copyOf(buf, p);
}
void tr(Object... o) { if(INPUT.length() > 0)System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: [n,k]=[int(j) for j in input().split()]
a=0
while n%2==0:
a+=1
n=n//2
if k>a:
print((k-a-1)//4+1)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,k;
cin>>n>>k;
while(k&&n%2==0){
n/=2;
--k;
}
cout<<(k+3)/4;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: function supermarket(prices, n, k) {
// write code here
// do not console.log the answer
// return sorted array
const newPrices = prices.sort((a, b) => a - b).slice(2)
let kk = k
let price = 0;
let i = 0
while (kk-- && i < newPrices.length) {
price += newPrices[i]
i += 1
}
return price
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str[]= br.readLine().trim().split(" ");
int n=Integer.parseInt(str[0]);
int k=Integer.parseInt(str[1]);
str= br.readLine().trim().split(" ");
int arr[]=new int[n];
for(int i=0;i<n;i++)
arr[i]=Integer.parseInt(str[i]);
Arrays.sort(arr);
long sum=0;
for(int i=2;i<k+2;i++)
sum+=arr[i];
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: a=input().split()
for i in range(len(a)):
a[i]=int(a[i])
b=input().split()
for i in range(len(b)):
b[i]=int(b[i])
b.sort()
b.reverse()
b.pop()
b.pop()
s=0
while a[1]>0:
s+=b.pop()
a[1]-=1
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n, k;
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
int ans = 0;
for(int i = 3; i <= 2 + k; i++)
ans += a[i];
cout << ans << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a rook is placed at a position (X, Y). Your task is to find the minimum steps Rook will take to go to the position (1, 1).The input contains two integer X and Y.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Print the number of steps rook will take to go to the position (X, Y).Sample Input:-
2 4
Sample Output:-
2
Explanation:-
one of the possible paths is:-
(2, 4) - > (2, 1) - > (1, 1)
Sample Input:-
1 2
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
int X = scan.nextInt();
int Y = scan.nextInt();
if((X == 1) && (Y == 1)){
System.out.println(0);
}
else if((X == 1) || (Y == 1)){
System.out.println(1);
}
else{
System.out.println(2);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a rook is placed at a position (X, Y). Your task is to find the minimum steps Rook will take to go to the position (1, 1).The input contains two integer X and Y.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Print the number of steps rook will take to go to the position (X, Y).Sample Input:-
2 4
Sample Output:-
2
Explanation:-
one of the possible paths is:-
(2, 4) - > (2, 1) - > (1, 1)
Sample Input:-
1 2
Sample Output:-
1, I have written this Solution Code: X, Y = [int(x) for x in input().split()]
if X == 1 and Y == 1:
print(0)
elif X == 1 or Y == 1:
print(1)
else:
print(2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a rook is placed at a position (X, Y). Your task is to find the minimum steps Rook will take to go to the position (1, 1).The input contains two integer X and Y.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Print the number of steps rook will take to go to the position (X, Y).Sample Input:-
2 4
Sample Output:-
2
Explanation:-
one of the possible paths is:-
(2, 4) - > (2, 1) - > (1, 1)
Sample Input:-
1 2
Sample Output:-
1, I have written this Solution Code: #include <iostream>
using namespace std;
int Rook(int X, int Y){
//Enter your code here
if(X==1 && Y==1){return 0;}
if(X==1 || Y==1){return 1;}
return 2;
}
int main(){
int x,y;
scanf("%d%d",&x,&y);
printf("%d",Rook(x,y));
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton has shifted to a new house and it came to his notice that there is only one socket in his house !!!
Newton does a lot of experiments using electricity and wants at least B sockets in his house.
Furthermore, he has an ample amount of extension cords and each extension has A sockets and takes one socket. In others words, an extension cord extends one socket into A sockets. You need to tell Newton how many extension cords will be required so that he can have at least B sockets.The first and the only line of the input contains 2 single integers A and B
<b>Constraints:</b>
1) 2 ≤ A ≤ 200
2) 1 ≤ B ≤ 200Print the answerSample Input 1:
4 10
Sample Output 1:
3
Sample Input 2:
8 9
Sample Output 2:
2
<b>Explanation 1:</b>
3 power strips, each with 4 sockets, extend the socket into 10 empty sockets.
, I have written this Solution Code: ///#include<bits/stdc++.h>
#include<iostream>
using namespace std;
#include<cmath>
#include<cstdio>
#include<cctype>
#include<vector>
#include<bitset>
#include<map>
#include<set>
#include<algorithm>
#include<cstdio>
#define pi acos(-1.0)
#define nl "\n"
/*data type ------------------------------------*/
#define ulli unsigned long long int
#define lli long long int
#define ld long double
/*---------------------------------------------*/
/*container -----------------------------------*/
#define vint vector<int>
#define vll vector<long long>
#define vlli vector<long long int>
#define vchar vector<char>
/*---------------------------------------------*/
/*container input and output-------------------*/
#define inv(v) for(auto& i:v) cin>>i
#define outv(v) for(auto& i:v) cout<<i<<" "
/*---------------------------------------------*/
/*container function --------------------------*/
#define sum(a) ( accumulate ((a).begin(), (a).end(), 0ll))
#define mine(a) (*min_element((a).begin(), (a).end()))
#define maxe(a) (*max_element((a).begin(), (a).end()))
#define mini(a) ( min_element((a).begin(), (a).end()) - (a).begin())
#define maxi(a) ( max_element((a).begin(), (a).end()) - (a).begin())
#define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define Sort(a) ( sort((a).begin(), (a).end()))
#define Sort_d(a) ( sort(a.begin(), a.end(), greater<int>()))
/*---------------------------------------------*/
#define precision(n) fixed<<setprecision(n)
#define fast ios_base::sync_with_stdio(0);cin.tie(0); cout.tie(0);
/*my function ------------------------------------*/
#define max2(x,y) ((x>y)?x:y)
#define min2(x,y) ((x<y)?x:y)
#define lcm(x,y) ((x*y)/(__gcd(x,y)))
#define mod 1000000007
/*bit manipulation---------------------------------*/
#define twopower(n) (1<<n)
#define MAX INT_MAX
#define MIN INT_MIN
#define ishigh(num, position) ((num&(1<<position))>0? 1:0)
#define ll_count_high(num) (__builtin_popcountll(num))
#define int_count_high(num) (__builtin_popcount(num))
#define toupper(char) (char & '_')
//cout<<(char)toupper('a')<<nl; out:A
#define tolower(char) (char | ' ')
//cout<<(char)tolower('B')<<nl; out:b
/*-------------------------------------------------*/
/*string convert_int_to_string(int number)
{
stringstream convert;
convert<<number;
string s = convert.str();
return s;
}*/
//find big factorial using python-3
/*
#User function Template for python3
import math
class Solution:
def factorial(self, N):
x = math.factorial(N)
list = []
s = str(x)
for j in s:
list.append(int(j))
return list
#{
# Driver Code Starts
#Initial Template for Python 3
if __name__ == '__main__':
t=int(input())
for _ in range(t):
N = int(input())
ob = Solution()
ans = ob.factorial(N);
for i in ans:
print(i,end="")
print()
*/
/*vector<lli>All_perfect_square(lli range)
{
vector<lli>perfect_square;
for(lli x = 0; x*x<=range; x++)
{
perfect_square.push_back(x*x);
}
return perfect_square;
}*/
/*lli find_near_perfect_square(vector<lli>&vect, lli number)
{
//use binary search
lli ans = -1;
lli left = 0, right = vect.size()-1, mid = (left+right)/2;
while(left<=right)
{
if(vect[mid]>=number){ans = vect[mid];right = mid-1; mid = (left+right)/2;}
else {left = mid+1;mid = (left+right)/2;}
}
return ans;
}*/
//sieve algorithm
/*vector<int>all_prime;
long long int index_array[5000] = {0};
void seive(int N)
{
index_array[0] = 1;
index_array[1] = 1;
all_prime.push_back(2);
for(long long int i = 3; i<=N; i+=2)
{
if(index_array[i]==0)
{
all_prime.push_back(i);
for(long long int j = i*i; j<=N; j+=i*2)
{
index_array[j] = 1;
}
}
}
}
*/
//find x power y using bit manipulation. complexity O(log N)
/*int bitwise_pow(int a, int n)
{
int ans = 1;// Stores final answer
while (n > 0) {
bool last_bit = (n & 1);
// Check if current LSB is set
if (last_bit){ans = ans * a;}
a = a * a;
n = n >> 1;// Right shift
}
return ans;
}
*/
/*bool isPrime(int n)
{
if (n <= 1)
return false;
// Check from 2 to square root of n
for (int i = 2; i <= sqrt(n); i++)
if (n % i == 0)
return false;
return true;
}
lli N = 100000;
*/
/*vint ar;
vint ar2;
int p = 1;
int check(int n)
{
//cout<<n<<endl;
int sz = ar.size(), sz2 = ar2.size(), no = 0;
for(int i = p; i<n; i*=n)
{
cout<<"i = "<<i<<nl;
no = 0;
for(int j = 0; j<sz; j++)
{
if(ar[j]%i==0){no = 1;break;}
}
if(n%i==0&&no==0)return i;
}
return 0;
}
void solve_the_problem()
{
int n, cnt = 0, ans = 0, x;cin>>n;
string s;cin>>s;
for(int i = 1; i<=n; i++)
{
if(s[i-1]=='0')
{
x = check(i);
//cout<<x<<nl;
if(x==0)
{
ans+=i;
}
else ans+=x;
cout<<ans<<nl;
}
else ar.push_back(i);
}
cout<<ans<<nl;
ar.clear();
p = 1;
// cout<<"......"<<endl;
// outv(ar);
// cout<<nl;
}
*/
void solve_the_problem(int T)
{
int a, b, cnt = 0;
cin>>a>>b;
int soket = 1;
while(soket<b)
{
soket--;
soket+=a;
cnt++;
}
cout<<cnt<<nl;
}
int main()
{
//int t; cin>>t;
int t = 1;
for(int i = 1; i<=t; i++) solve_the_problem(t);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers, You have to print the total number of non-empty subarray count of an array.The first line of the input contains a single integer N denoting the size of an array.
The second line of the input contains space-separated integers A<sub>1</sub>, A<sub>2</sub>,. , A<sub>n</sub>.
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ A[i] ≤ 10<sup>5</sup>
<b>Note</b>: Use long for big multiplications.Print the count of the total number of subarrays in an array.Sample Input
3
1 2 3
Sample Output
6
Explanation
[1], [2], [3], [1, 2], [2, 3], [1, 2, 3]
Hence the total number of non empty subarrays are 6., I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
auto start = std::chrono::high_resolution_clock::now();
int tt = 1;
// cin >> tt;
while (tt--) {
long n;
cin >> n;
for (int i = 0; i < n; i++) {
int temp;
cin >> temp;
}
long ans = (n * (n + 1)) / 2;
cout << ans << "\n";
}
auto stop = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Cf cf = new Cf();
cf.solve();
}
static class Cf {
InputReader in = new InputReader(System.in);
OutputWriter out = new OutputWriter(System.out);
int mod = (int)1e9+7;
public void solve() {
int t = in.readInt();
if(t>=6) {
out.printLine("No");
}else {
out.printLine("Yes");
}
}
public long findPower(long x,long n) {
long ans = 1;
long nn = n;
while(nn>0) {
if(nn%2==1) {
ans = (ans*x) % mod;
nn-=1;
}else {
x = (x*x)%mod;
nn/=2;
}
}
return ans%mod;
}
public static int log2(int x) {
return (int) (Math.log(x) / Math.log(2));
}
private static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public double readDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
private static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
writer.flush();
}
public void printLine(Object... objects) {
print(objects);
writer.println();
writer.flush();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n<6)
cout<<"Yes";
else
cout<<"No";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: n=int(input())
if(n>=6):
print("No")
else:
print("Yes"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: static void simpleSum(int a, int b, int c){
System.out.println(a+b+c);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: void simpleSum(int a, int b, int c){
cout<<a+b+c;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: x = input()
a, b, c = x.split()
a = int(a)
b = int(b)
c = int(c)
print(a+b+c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",end=' ')
else:
print(i,end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a class <code>FooBar</code>, whose objects have <code>foo</code> and <code>bar</code> as their attributes.
Also, write the function <code>solve(obj, param1, param2)</code>.
<code>obj:</code> Object of class FooBar
<code>param1, param2:</code> some random numbers provided to you directly as input.
See the example for more clarity.Input will have <code>4</code> numbers, viz. <code>foo</code>,<code>bar</code>,<code>param1</code>, <code>param2</code> separated by spaces.
You need not worry about the same, it is handled properly by the hidden pre-function code.
Example:
4 5 6 7Function <code>solve(obj, param1, param2)</code> should return the value <code>obj.foo + obj.bar + func(param1, param2)</code>
There is a global function named <code>func(a, b)</code>, which takes two numbers as parameters, and its implementation is hidden.
<b>Note</b>: <code>func</code> uses properties of obj (using <code>this</code>, just like class methods), but does not accept obj as a parameter.const obj = new FooBar(1, 2); // say
console.log(solve(obj, param1, param2)) // prints a number as answer
// param1 and param2 are some hidden numbers passed as input.
, I have written this Solution Code: class FooBar {
constructor(foo, bar) {
this.foo = foo;
this.bar = bar;
}
}
function solve(obj, gfoo, gbar) {
const g = func.bind(obj);
return obj.foo + obj.bar + g(gfoo, gbar);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N. Print the third largest integer in the array in linear time.
Third largest element of the array is the third number in the array when sorted in non- increasing order.A single integer n.
Next line contains n integers separated by space.
<b>Constraints </b>
3 <= n <= 10<sup>5</sup>
-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup>A single integer denoting required answer.Input:
6
-2 7 3 1 6 5
Output:
5
Explanation :
sort in non- increasing order : 7 6 5 3 1 -2, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(in.next());
}
int x=-1,y=-1,z=-1;
for(int i=0;i<n;i++){
if(x==-1 || a[i] >= a[x]){
z=y;
y=x;
x=i;
}
else if(y==-1 || a[i] >= a[y]){
z=y;
y=i;
}
else if(z == -1 || a[i] >= a[z]){
z=i;
}
}
out.print(a[z]);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array <b>A</b> of size <b>N</b>. Find the minimum subsequence size of this array such that the sum of its elements is <b>at least K</b>. If there is no such subsequence possible, print -1.
<b>Subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. </b>First line of the input contains two integers N and K.
The second line of input contains N space- seperated integers.
Constraint:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>14</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the minimum size of the required subsequence. If there is no such possible subsequence, print -1.Sample Input:
5 17
9 4 7 6 2
Sample Output:
3
Explaination:
We can use the subsequence {4, 7, 6}. Their sum = 4 + 7 + 6 = 17.
We cannot obtain sum 17 in less than 3 subsequence size., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader r = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(r);
String line[] = br.readLine().strip().split(" ");
int n = Integer.parseInt(line[0]);
long k = Long.parseLong(line[1]);
int arr[] = new int[n];
line = br.readLine().strip().split(" ");
for(int i=0; i<n; i++){
arr[i] = Integer.parseInt(line[i]);
}
Arrays.sort(arr);
long sum = 0;
int ans = -1;
for(int i=n-1; i>=0; i--){
sum+=arr[i];
if(sum >= k){
ans = n-1-i+1;
break;
}
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array <b>A</b> of size <b>N</b>. Find the minimum subsequence size of this array such that the sum of its elements is <b>at least K</b>. If there is no such subsequence possible, print -1.
<b>Subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. </b>First line of the input contains two integers N and K.
The second line of input contains N space- seperated integers.
Constraint:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>14</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the minimum size of the required subsequence. If there is no such possible subsequence, print -1.Sample Input:
5 17
9 4 7 6 2
Sample Output:
3
Explaination:
We can use the subsequence {4, 7, 6}. Their sum = 4 + 7 + 6 = 17.
We cannot obtain sum 17 in less than 3 subsequence size., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long long ll;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve() {
int n, k;
cin >> n >> k;
vector<int> a(n);
for(auto &i : a) cin >> i;
sort(rall(a));
int cur = 0;
for(int i = 0; i < n; i++){
cur += a[i];
if(cur >= k){
cout << i + 1;
return;
}
}
cout << -1;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine().trim();
char n[] = s.toCharArray();
int posMin = 0;
for(int i=0;i<n.length;i++){
if(n[posMin]>n[i])
posMin = i;
}
for(int i=0;i<n.length;i++)
n[i] = n[posMin];
System.out.print(String.valueOf(n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
char c='z';
for(auto r:s)
c=min(c,r);
for(int i=0;i<s.length();++i)
cout<<c;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: s=input()
l=len(s)
k=min(s)
m=k*l
print(m), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two binary trees, the task is to find if both of them are identical or not.
Note: Here identical means exactly same.User task:
Since this is a functional problem you don't have to worry about input, you just have to complete the function <b>isIdentical()</b> that takes the root of both the trees as input.
Constraints:
1 <= T <= 10
1 <= N <= 100
1 <= Data of Nodes <= 100
For <b>Custom Input:</b>
The first line of input should contain the number of test cases T. For each test case, there will be two lines of input.
The first line contains a number of nodes N. Second line will be a string representing the tree as described below:
The values in the string are such that for every parent you need to mention its child and state whether its is left or right like- 1 2 L 1 3 RFor each test case you need to return true or false based on whether they are identical or not. If the trees are identical then driver code will print 1 otherwise 0Sample Input:
1
/ \
2 3
1
/ \
2 3
Sample Output:
1
Sample Input:-
1
/ \
2 3
1
/ \
3 2
Sample Output:-
0
Explanation:
Test case 1: There are two trees both having 3 nodes and 2 edges, both trees are identical having the root as 1, left child of 1 is 2 and right child of 1 is 3.
Test case 2: There are two trees both having 3 nodes and 2 edges, but both trees are not identical, I have written this Solution Code: static boolean isIdentical(Node root1, Node root2)
{
if (root1 == null && root2 == null) {
return true;
}
if (root1 == null || root2 == null) {
return false;
}
if (root1.data != root2.data) {
return false;
}
return isIdentical(root1.left, root2.left) && isIdentical(root1.right, root2.right);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: static void pattern(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j + " ");
}
System.out.println();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
printf("\n");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: function pattern(n) {
// write code herenum
for(let i = 1;i<=n;i++){
let str = ''
for(let k = 1; k <= i;k++){
if(k === 1) {
str += `${k}`
}else{
str += ` ${k}`
}
}
console.log(str)
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
printf("\n");
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: def patternPrinting(n):
for i in range(1,n+1):
for j in range (1,i+1):
print(j,end=' ')
print()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array Arr of n integers. You can perform the following operations on the array any number of times.
Operation 1: Add or Subtract 2 from any element of the array any number of times
Operation 2: Remove a number from the array
Your aim is to make all the elements of the array equal after performing the above operations any number of times. Report the maximum size of the array possible.The first line of input contains N, the size of the array
The second line of input contains N space-separated integers
Constraints
1 ≤ N ≤ 100000
1 ≤ Arr[i] ≤ 1000000000The output should contain only one integer, the maximum size of the array possible such that all elements of the array are equal.Sample Input 1
5
1 2 3 3 2
Sample Output 1
3
Sample Input 2
2
1 2
Sample Output 2
1
Explanation:-
Testcase1 :- you can remove both 2 from the array making the array equal to 1,3,3. Now subtract 2 from both 3 making the array equal 1,1,1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
int size = Integer.parseInt(bf.readLine());
String s[] = bf.readLine().split(" ");
int countEven =0;
int countOdd = 0;
for(int i=0;i<size;i++){
if(Integer.parseInt(s[i])%2 == 0){
countEven++;
}else{
countOdd++;
}
}
System.out.println(Math.max(countOdd,countEven));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array Arr of n integers. You can perform the following operations on the array any number of times.
Operation 1: Add or Subtract 2 from any element of the array any number of times
Operation 2: Remove a number from the array
Your aim is to make all the elements of the array equal after performing the above operations any number of times. Report the maximum size of the array possible.The first line of input contains N, the size of the array
The second line of input contains N space-separated integers
Constraints
1 ≤ N ≤ 100000
1 ≤ Arr[i] ≤ 1000000000The output should contain only one integer, the maximum size of the array possible such that all elements of the array are equal.Sample Input 1
5
1 2 3 3 2
Sample Output 1
3
Sample Input 2
2
1 2
Sample Output 2
1
Explanation:-
Testcase1 :- you can remove both 2 from the array making the array equal to 1,3,3. Now subtract 2 from both 3 making the array equal 1,1,1, I have written this Solution Code: n=int(input())
l=list(map(int,input().split()))
c=0
for i in l:
if(i%2==0):
c+=1
if(n-c>c):
print(abs(n-c))
else:
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array Arr of n integers. You can perform the following operations on the array any number of times.
Operation 1: Add or Subtract 2 from any element of the array any number of times
Operation 2: Remove a number from the array
Your aim is to make all the elements of the array equal after performing the above operations any number of times. Report the maximum size of the array possible.The first line of input contains N, the size of the array
The second line of input contains N space-separated integers
Constraints
1 ≤ N ≤ 100000
1 ≤ Arr[i] ≤ 1000000000The output should contain only one integer, the maximum size of the array possible such that all elements of the array are equal.Sample Input 1
5
1 2 3 3 2
Sample Output 1
3
Sample Input 2
2
1 2
Sample Output 2
1
Explanation:-
Testcase1 :- you can remove both 2 from the array making the array equal to 1,3,3. Now subtract 2 from both 3 making the array equal 1,1,1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
long long a;
int cnt;
for(int i=0;i<n;i++){
cin>>a;
if(a&1){cnt++;}
}
cout<<max(cnt,n-cnt);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args)throws IOException {
Reader sc = new Reader();
int N = sc.nextInt();
int[] arr = new int[N];
for(int i=0;i<N;i++){
arr[i] = sc.nextInt();
}
int max=0;
if(arr[0]<arr[N-1])
System.out.print(N-1);
else{
for(int i=0;i<N-1;i++){
int j = N-1;
while(j>i){
if(arr[i]<arr[j]){
if(max<j-i){
max = j-i;
} break;
}
j--;
}
if(i==j)
break;
if(j==N-1)
break;
}
if(max==0)
System.out.print("-1");
else
System.out.print(max);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
/* For a given array arr[],
returns the maximum j – i such that
arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
int maxDiff;
int i, j;
int *LMin = new int[(sizeof(int) * n)];
int *RMax = new int[(sizeof(int) * n)];
/* Construct LMin[] such that
LMin[i] stores the minimum value
from (arr[0], arr[1], ... arr[i]) */
LMin[0] = arr[0];
for (i = 1; i < n; ++i)
LMin[i] = min(arr[i], LMin[i - 1]);
/* Construct RMax[] such that
RMax[j] stores the maximum value from
(arr[j], arr[j+1], ..arr[n-1]) */
RMax[n - 1] = arr[n - 1];
for (j = n - 2; j >= 0; --j)
RMax[j] = max(arr[j], RMax[j + 1]);
/* Traverse both arrays from left to right
to find optimum j - i. This process is similar to
merge() of MergeSort */
i = 0, j = 0, maxDiff = -1;
while (j < n && i < n)
{
if (LMin[i] < RMax[j])
{
maxDiff = max(maxDiff, j - i);
j = j + 1;
}
else
i = i + 1;
}
return maxDiff;
}
// Driver Code
signed main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int maxDiff = maxIndexDiff(a, n);
cout << maxDiff;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
rightMax = [0] * n
rightMax[n - 1] = arr[n - 1]
for i in range(n - 2, -1, -1):
rightMax[i] = max(rightMax[i + 1], arr[i])
maxDist = -2**31
i = 0
j = 0
while (i < n and j < n):
if (rightMax[j] >= arr[i]):
maxDist = max(maxDist, j - i)
j += 1
else:
i += 1
if maxDist==0:
maxDist=-1
print(maxDist), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array, your task is to sort the array using merge sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>implementMergeSort()</b> that takes 3 arguments.
arr: input array
start: starting index which is 0
end: ending index of array
Constraints
1 <= T <= 100
1 <= N <= 10<sup>6</sup>
0 <= Arr[i] <= 10<sup>9</sup>
Sum of 'N' over all test cases does not exceed 10<sup>6</sup>You need to return the sorted array. The driver code will print the array in sorted form.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] implementMergeSort(int arr[], int start, int end)
{
if (start < end)
{
// Find the middle point
int mid = (start+end)/2;
// Sort first and second halves
implementMergeSort(arr, start, mid);
implementMergeSort(arr , mid+1, end);
// Merge the sorted halves
merge(arr, start, mid, end);
}
return arr;
}
public static void merge(int arr[], int start, int mid, int end)
{
// Find sizes of two subarrays to be merged
int n1 = mid - start + 1;
int n2 = end - mid;
/* Create temp arrays */
int L[] = new int [n1];
int R[] = new int [n2];
/*Copy data to temp arrays*/
for (int i=0; i<n1; ++i)
L[i] = arr[start + i];
for (int j=0; j<n2; ++j)
R[j] = arr[mid + 1+ j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = start;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array, your task is to sort the array using merge sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>implementMergeSort()</b> that takes 3 arguments.
arr: input array
start: starting index which is 0
end: ending index of array
Constraints
1 <= T <= 100
1 <= N <= 10<sup>6</sup>
0 <= Arr[i] <= 10<sup>9</sup>
Sum of 'N' over all test cases does not exceed 10<sup>6</sup>You need to return the sorted array. The driver code will print the array in sorted form.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: for _ in range(int(input())):
n = int(input())
print(*sorted(list(map(int,input().split())))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Consider the integer sequence A = {1, 2, 3, ...., N} i.e. the first N natural numbers in order.
You are now given two integers, L and S. Determine whether there exists a subarray with length L and sum S after removing <b>at most one</b> element from A.
A <b>subarray</b> of an array is a non-empty sequence obtained by removing zero or more elements from the front of the array, and zero or more elements from the back of the array.The first line contains a single integer T, the number of test cases.
T lines follow. Each line describes a single test case and contains three integers: N, L, and S.
<b>Constraints:</b>
1 <= T <= 100
2 <= N <= 10<sup>9</sup>
1 <= L <= N-1
1 <= S <= 10<sup>18</sup> <b> (Note that S will not fit in a 32-bit integer) </b>For each testcase, print "YES" (without quotes) if a required subarray can exist, and "NO" (without quotes) otherwise.Sample Input:
3
5 3 11
5 3 5
5 3 6
Sample Output:
YES
NO
YES
Sample Explanation:
For the first test case, we can remove 3 from A to obtain A = {1, 2, 4, 5} where {2, 4, 5} is a required subarray of size 3 and sum 11., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long INF = 1e18;
const int INFINT = INT_MAX/2;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x, y) freopen(x, "r", stdin); freopen(y, "w", stdout);
#define out(x) cout << ((x) ? "YES\n" : "NO\n")
#define CASE(x, y) cout << "Case #" << x << ":" << " \n"[y]
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now(); auto end_time = start_time;
#define measure() end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
#define reset_clock() start_time = std::chrono::high_resolution_clock::now();
typedef long long ll;
typedef long double ld;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll norm(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; g--; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n): adj(n+1) {}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
struct LCA
{
vll tin, tout, level;
vector<vll> up;
ll timer;
void _dfs(vector<vector<int>> &adj, ll x, ll par=-1)
{
up[x][0] = par;
REP(i, 1, 20)
{
if(up[x][i-1] == -1) break;
up[x][i] = up[up[x][i-1]][i-1];
}
tin[x] = ++timer;
for(auto &p: adj[x])
{
if(p != par)
{
level[p] = level[x]+1; _dfs(adj, p, x);
}
}
tout[x] = ++timer;
}
LCA(Graph &G, ll root)
{
int n = G.adj.size();
tin.resize(n); tout.resize(n); up.resize(n, vll(20, -1)); level.resize(n, 0);
timer = 0;
//does not handle forests, easy to modify to handle
_dfs(G.adj, root);
}
ll find_kth(ll x, ll k)
{
ll cur = x;
REP(i, 0, 20)
{
if((k>>i)&1) cur = up[cur][i];
if(cur == -1) break;
}
return cur;
}
bool is_ancestor(ll x, ll y)
{
return tin[x]<=tin[y]&&tout[x]>=tout[y];
}
ll find_lca(ll x, ll y)
{
if(is_ancestor(x, y)) return x;
if(is_ancestor(y, x)) return y;
ll best = x;
REPd(i, 19, 0)
{
if(up[x][i] == -1) continue;
else if(is_ancestor(up[x][i], y)) best = up[x][i];
else x = up[x][i];
}
return best;
}
ll dist(ll a, ll b)
{
return level[a] + level[b] - 2*level[find_lca(a, b)];
}
};
long long ap_sum(long long st, long long len) {
//diff is always 1
long long en = st + len - 1;
return (len * (st + en))/2;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
for(int i=0; i<t; i++) {
long long N, L, S;
cin >> N >> L >> S;
if(S < ap_sum(1, L) || S > ap_sum(N - L + 1, L)) {
cout << "NO\n";
} else {
cout << "YES\n";
}
}
return 0;
}
/*
*/, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: def get_opposite_face(n):
return 7-n
t = int(input())
for n in range(t):
print(get_opposite_face(int(input()))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
cout << 7-n << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(6-n+1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four positive integers A, B, C, and D, can a rectangle have these integers as its sides?The input consists of a single line containing four space-separated integers A, B, C and D.
<b> Constraints: </b>
1 ≤ A, B, C, D ≤ 1000Output "Yes" if a rectangle is possible; otherwise, "No" (without quotes).Sample Input 1:
3 4 4 3
Sample Output 1:
Yes
Sample Explanation 1:
A rectangle is possible with 3, 4, 3, and 4 as sides in clockwise order.
Sample Input 2:
3 4 3 5
Sample Output 2:
No
Sample Explanation 2:
No rectangle can be made with these side lengths., I have written this Solution Code: #include<iostream>
using namespace std;
int main(){
int a,b,c,d;
cin >> a >> b >> c >> d;
if((a==c) &&(b==d)){
cout << "Yes\n";
}else if((a==b) && (c==d)){
cout << "Yes\n";
}else if((a==d) && (b==c)){
cout << "Yes\n";
}else{
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: During the elections, Bob is in charge of conducting voting in his village, but the EVM system malfunctioned, and there was a long line of voters waiting outside to vote. The following is how the Advanced EVM Machine works.
Each time when a voter scans his VoterId Card and votes for the party of his choice, the Voter's id and Party Name are registered in the background, and if the same voter votes again, the EVM does not capture his vote, then the vote is skipped and the vote given the first time is used.
Now that you are Bob's best mate, you can't bear to see him in such a strained situation when outside voters are being very aggressive and screaming at him. Can you easily write a piece of code to save your friend's life while Bob is busy calming down the outside situation?The number N (1 ≤ N ≤ 1e5) appears on the first line. The queries to the machine are included in the next n lines. Each request consists of two strings and is written on a non-empty line. The first string is a Voter Card Id, and the second string is the Party Name, all of which are at most 32 characters long both upper case and lower case possible.
<b>Constraints</b>
1 ≤ N ≤ 100000
1 ≤ Voter Id length ≤ 40
1 ≤ PartyName length ≤ 32Output a single line indicating which party has won and by how many votes. In case of a draw between parties print all the parties with the same winning vote, in lexographically increasing order on basis of Party Name.Sample Input
4
12678345 BJP
57891082 congress
12678345 AAP
65489 TMC
Sample Output
BJP 1
Congress 1
TMC 1
<b>Explanation :</b>
As Winning Parties here as BJP, Congress, TMC with 1 vote(s) each, but AAP vote is not considered because the same VoterId - 12678345 has done a vote again., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
HashSet<String> set = new HashSet<String> ();
HashMap<String, Long> map = new HashMap<>();
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
long max = 1L;
for(int i=0;i<n;i++){
String[] num = br.readLine().split(" ");
String voterID = num[0];
String partyName = num[1];
if(!set.contains(voterID)){
set.add(voterID);
if(!map.containsKey(partyName)){
map.put(partyName, 1L);
}
else{
map.put(partyName, map.get(partyName)+1);
max = Math.max(max, map.get(partyName));
}
}
}
ArrayList<String> winningParty = new ArrayList<>();
for (Map.Entry<String,Long> entry : map.entrySet()){
if(entry.getValue()==max) winningParty.add(entry.getKey());
}
Collections.sort(winningParty);
for(int i=0;i<winningParty.size();i++){
System.out.println(winningParty.get(i)+" "+max);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: During the elections, Bob is in charge of conducting voting in his village, but the EVM system malfunctioned, and there was a long line of voters waiting outside to vote. The following is how the Advanced EVM Machine works.
Each time when a voter scans his VoterId Card and votes for the party of his choice, the Voter's id and Party Name are registered in the background, and if the same voter votes again, the EVM does not capture his vote, then the vote is skipped and the vote given the first time is used.
Now that you are Bob's best mate, you can't bear to see him in such a strained situation when outside voters are being very aggressive and screaming at him. Can you easily write a piece of code to save your friend's life while Bob is busy calming down the outside situation?The number N (1 ≤ N ≤ 1e5) appears on the first line. The queries to the machine are included in the next n lines. Each request consists of two strings and is written on a non-empty line. The first string is a Voter Card Id, and the second string is the Party Name, all of which are at most 32 characters long both upper case and lower case possible.
<b>Constraints</b>
1 ≤ N ≤ 100000
1 ≤ Voter Id length ≤ 40
1 ≤ PartyName length ≤ 32Output a single line indicating which party has won and by how many votes. In case of a draw between parties print all the parties with the same winning vote, in lexographically increasing order on basis of Party Name.Sample Input
4
12678345 BJP
57891082 congress
12678345 AAP
65489 TMC
Sample Output
BJP 1
Congress 1
TMC 1
<b>Explanation :</b>
As Winning Parties here as BJP, Congress, TMC with 1 vote(s) each, but AAP vote is not considered because the same VoterId - 12678345 has done a vote again., I have written this Solution Code: /**
* author: tourist1256
* created: 2021-4-2 13:58:11
**/
#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <vector>
using namespace std;
double pi = acos(-1);
#define tezi \
ios_base::sync_with_stdio(false); \
cin.tie(0); \
cout.tie(0);
#define F_OR(i, a, b, s) \
for (ll i = (a); (s) > 0 ? i < (b) : i >= (b); i += (s))
#define F_OR1(e) F_OR(i, 0, e, 1)
#define F_OR2(i, e) F_OR(i, 0, e, 1)
#define F_OR3(i, b, e) F_OR(i, b, e, 1)
#define F_OR4(i, b, e, s) F_OR(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)
#define FOR(...) \
F_ORC(__VA_ARGS__) \
(__VA_ARGS__)
#define EACH(x, a) for (auto &x : a)
#define itr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define fill(a, b) memset(a, b, sizeof(a))
#define LOCAL
#define time cout << (0.1 * clock()) / CLOCKS_PER_SEC << endl;
#define countBits(x) __builtin_popcount(ll(x))
#define countZeroesAtBegin(x) __builtin_clz(ll(x))
#define countZeroesAtEnd(x) __builtin_ctz(ll(x))
#define last(x) x[x.end() - x.begin() - 1]
#define pb push_back
#define bg begin
#define ff first
#define ss second
#define pi 3.1415926535897932384626
#define infll 0x3f3f3f3f3f3f3f3f
using ll = long long;
using pl = pair<ll, ll>;
using vll = vector<ll>;
using vpl = vector<pl>;
using mat = vector<vll>;
const ll MAX = 1e5;
const ll mod = 1e9 + 7;
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
static int row[4] = {-1, 0, 0, 1};
static int col[4] = {0, -1, 1, 0};
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
EACH(ele, v) { cout << ele << " "; }
return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
return os << '(' << p.ff << "," << p.ss << ')';
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
EACH(ele, v) { cin >> ele; }
return in;
}
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
void _runtime_terror_() {
ll N, winningVotes = -infll;
cin >> N;
string voterId, partyName;
map<string, ll> database;
map<string, bool> voters;
for (int i = 0; i < N; i++) {
cin >> voterId >> partyName;
if (voters.find(voterId) == voters.end()) {
voters[voterId] = true;
database[partyName] += 1;
winningVotes = max(database[partyName], winningVotes);
}
}
vector<pair<string, ll>> ans;
for (auto &it : database) {
if (winningVotes == it.ss) {
ans.pb({it.ff, it.ss});
}
}
for (auto &it : ans) {
cout << it.ff << " " << it.ss << "\n";
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
freopen("debug.txt", "w", stderr);
#endif
tezi;
int tc = 1;
// cin >> tc;
while (tc--) {
_runtime_terror_();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Phoebe is a big GCD fan. Being bored, she starts counting number of pairs of integers (A, B) such that following conditions are satisfied:
<li> GCD(A, B) = X (As X is Phoebe's favourite integer)
<li> A <= B <= L
As Phoebe's performance is coming up, she needs your help to find the number of such pairs possible.
Note: GCD refers to the <a href="https://en.wikipedia.org/wiki/Greatest_common_divisor">Greatest common divisor</a>.Input contains two integers L and X.
Constraints:
1 <= L, X <= 1000000000Print a single integer denoting number of pairs possible.Sample Input
5 2
Sample Output
2
Explanation: Pairs satisfying all conditions are: (2, 2), (2, 4)
Sample Input
5 3
Sample Output
1
Explanation: Pairs satisfying all conditions are: (3, 3), I have written this Solution Code: import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve()
{
out.println(sumTotient(ni()/ni()));
}
public static int[] enumTotientByLpf(int n, int[] lpf)
{
int[] ret = new int[n+1];
ret[1] = 1;
for(int i = 2;i <= n;i++){
int j = i/lpf[i];
if(lpf[j] != lpf[i]){
ret[i] = ret[j] * (lpf[i]-1);
}else{
ret[i] = ret[j] * lpf[i];
}
}
return ret;
}
public static int[] enumLowestPrimeFactors(int n)
{
int tot = 0;
int[] lpf = new int[n+1];
int u = n+32;
double lu = Math.log(u);
int[] primes = new int[(int)(u/lu+u/lu/lu*1.5)];
for(int i = 2;i <= n;i++)lpf[i] = i;
for(int p = 2;p <= n;p++){
if(lpf[p] == p)primes[tot++] = p;
int tmp;
for(int i = 0;i < tot && primes[i] <= lpf[p] && (tmp = primes[i]*p) <= n;i++){
lpf[tmp] = primes[i];
}
}
return lpf;
}
public static long sumTotient(int n)
{
if(n == 0)return 0L;
if(n == 1)return 1L;
int s = (int)Math.sqrt(n);
long[] cacheu = new long[n/s];
long[] cachel = new long[s+1];
int X = (int)Math.pow(n, 0.66);
int[] lpf = enumLowestPrimeFactors(X);
int[] tot = enumTotientByLpf(X, lpf);
long sum = 0;
int p = cacheu.length-1;
for(int i = 1;i <= X;i++){
sum += tot[i];
if(i <= s){
cachel[i] = sum;
}else if(p > 0 && i == n/p){
cacheu[p] = sum;
p--;
}
}
for(int i = p;i >= 1;i--){
int x = n/i;
long all = (long)x*(x+1)/2;
int ls = (int)Math.sqrt(x);
for(int j = 2;x/j > ls;j++){
long lval = i*j < cacheu.length ? cacheu[i*j] : cachel[x/j];
all -= lval;
}
for(int v = ls;v >= 1;v--){
long w = x/v-x/(v+1);
all -= cachel[v]*w;
}
cacheu[(int)i] = all;
}
return cacheu[1];
}
void run() throws Exception
{
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
public static void main(String[] args) throws Exception { new Main().run(); }
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
private int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private double nd() { return Double.parseDouble(ns()); }
private char nc() { return (char)skip(); }
private String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}
private int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Phoebe is a big GCD fan. Being bored, she starts counting number of pairs of integers (A, B) such that following conditions are satisfied:
<li> GCD(A, B) = X (As X is Phoebe's favourite integer)
<li> A <= B <= L
As Phoebe's performance is coming up, she needs your help to find the number of such pairs possible.
Note: GCD refers to the <a href="https://en.wikipedia.org/wiki/Greatest_common_divisor">Greatest common divisor</a>.Input contains two integers L and X.
Constraints:
1 <= L, X <= 1000000000Print a single integer denoting number of pairs possible.Sample Input
5 2
Sample Output
2
Explanation: Pairs satisfying all conditions are: (2, 2), (2, 4)
Sample Input
5 3
Sample Output
1
Explanation: Pairs satisfying all conditions are: (3, 3), I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
ull X[20000001];
ull cmp(ull N){
return N*(N+1)/2;
}
ull solve(ull N){
if(N==1)
return 1;
if(N < 20000001 && X[N] != 0)
return X[N];
ull res = 0;
ull q = floor(sqrt(N));
for(int k=2;k<N/q+1;++k){
res += solve(N/k);
}
for(int m=1;m<q;++m){
res += (N/m - N/(m+1)) * solve(m);
}
res = cmp(N) - res;
if(N < 20000001)
X[N] = res;
return res;
}
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int l,x;
cin>>l>>x;
if(l<x)
cout<<0;
else
cout<<solve(l/x);
#ifdef ANIKET_GOYAL
cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bob has a string S consisting of lowercase English letters. He defines S′ to be the string after removing all "a" characters from S (keeping all other characters in the same order). He then generates a new string T by concatenating S and S′. In other words, T=S+S′.
You are given a string T. Your task is to find some S that Bob could have used to generate T. It can be shown that if an answer exists, it will be unique.The first line contains a single string S.
Constraints:
1<=|S|<=100000Print a string S that could have generated T. It can be shown if an answer exists, it is unique. If no string exists, print ":(" (without double quotes, there is no space between the characters).Sample Input 1:
ababacacbbcc
Sample Output 1:
ababacac
Explanations:
we have s= "ababacac", and s′= "bbcc", and t=s+s′= "ababacacbbcc"., I have written this Solution Code: import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String S = in.next();
StringBuilder sb = new StringBuilder();
for (char c : S.toCharArray()) {
if (c != 'a') {
sb.append(c);
}
}
boolean ok = true;
if (sb.length() % 2 == 0) {
int length = sb.length()/2;
for (int i=0; i<length; i++) {
if (sb.charAt(i) != sb.charAt(i+length)) {
ok = false;
break;
}
}
if (S.lastIndexOf('a') >= S.length() - length) {
ok = false;
}
if (ok) {
System.out.println(S.subSequence(0, S.length() - length));
}
} else {
ok = false;
}
if (!ok) {
System.out.println(":(");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bob has a string S consisting of lowercase English letters. He defines S′ to be the string after removing all "a" characters from S (keeping all other characters in the same order). He then generates a new string T by concatenating S and S′. In other words, T=S+S′.
You are given a string T. Your task is to find some S that Bob could have used to generate T. It can be shown that if an answer exists, it will be unique.The first line contains a single string S.
Constraints:
1<=|S|<=100000Print a string S that could have generated T. It can be shown if an answer exists, it is unique. If no string exists, print ":(" (without double quotes, there is no space between the characters).Sample Input 1:
ababacacbbcc
Sample Output 1:
ababacac
Explanations:
we have s= "ababacac", and s′= "bbcc", and t=s+s′= "ababacacbbcc"., I have written this Solution Code: inp = input()
inp1 = inp.replace('a','')
l = len(inp1)//2
ind = len(inp)-l
if(inp[0:ind]+inp1[l:] == inp ):
print(inp[0:ind])
else:
print(":("), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: n = int(input())
if (n%4==0 and n%100!=0 or n%400==0):
print("YES")
elif n==0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int n = scanner.nextInt();
LeapYear(n);
}
static void LeapYear(int year){
if(year%400==0 || (year%100 != 0 && year%4==0)){System.out.println("YES");}
else {
System.out.println("NO");}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's assume some functional definitions for this problem.
We take prime(x) as the set of all prime divisors of x. For example, prime(140)={2,5,7}, prime(169)={13}.
Let f(x,p) be the maximum possible integer p**k where k is an integer such that x is divisible by p**k.
(Here a**b means a raised to the power b or pow(a, b))
For example:
f(99,3)=9 (99 is divisible by 3**2=9 but not divisible by 3**3=27),
f(63,7)=7 (63 is divisible by 7**1=7 but not divisible by 7**2=49).
Let g(x,y) be the product of f(y,p) for all p in prime(x).
For example:
g(30,70)=f(70,2)*f(70,3)*f(70,5)=2*1*5=10,
g(525,63)=f(63,3)*f(63,5)*f(63,7)=9*1*7=63.
You are given two integers x and n. Calculate g(x,1)*g(x,2)*…*g(x,n) mod (1000000007).
(Read modulo exponentiation before attempting this problem)The only line contains integers x and n — the numbers used in formula.
Constraints
2 ≤ x ≤ 1000000000
1 ≤ n ≤ 1000000000000000000Print the answer corresponding to the input.Sample Input 1
10 2
Sample Output 1
2
Sample Input 2
20190929 1605
Sample Output 2
363165664
Explanation
In the first example, g(10,1)=f(1,2)⋅f(1,5)=1, g(10,2)=f(2,2)⋅f(2,5)=2., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static long mod =1000000007;
public static boolean isPrime(long m){
if (m <2)
return false;
for(int i =2 ;i*i<=m;i++)
{
if (m%i == 0)
return false;
}
return true;
}
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s [] =br.readLine().trim().split(" ");
long x = Long.parseLong(s[0]);
long n = Long.parseLong(s[1]);
long ans = 1;
for(int i = 2; i*i <= x; i++){
if(x%i != 0) continue;
ans = (ans*f(n, i)) % mod;
while(x%i == 0)
x /= i;
}
if(x > 1)
ans = (ans*f(n, x)) % mod;
System.out.println(ans);
}
static long f(long n, long p){
long ans = 1;
long cur = 1;
while(cur <= n/p){
cur = cur*p;
long z = power(p, n/cur);
ans = (ans*z) % mod;
}
return ans;
}
public static long power(long no,long pow){
long p = 1000000007;
long result = 1;
while(pow > 0)
{
if ( (pow & 1) == 1)
result = ((result%p) * (no%p))%p;
no = ((no%p) * (no%p))%p;
pow >>= 1;
}
return result;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's assume some functional definitions for this problem.
We take prime(x) as the set of all prime divisors of x. For example, prime(140)={2,5,7}, prime(169)={13}.
Let f(x,p) be the maximum possible integer p**k where k is an integer such that x is divisible by p**k.
(Here a**b means a raised to the power b or pow(a, b))
For example:
f(99,3)=9 (99 is divisible by 3**2=9 but not divisible by 3**3=27),
f(63,7)=7 (63 is divisible by 7**1=7 but not divisible by 7**2=49).
Let g(x,y) be the product of f(y,p) for all p in prime(x).
For example:
g(30,70)=f(70,2)*f(70,3)*f(70,5)=2*1*5=10,
g(525,63)=f(63,3)*f(63,5)*f(63,7)=9*1*7=63.
You are given two integers x and n. Calculate g(x,1)*g(x,2)*…*g(x,n) mod (1000000007).
(Read modulo exponentiation before attempting this problem)The only line contains integers x and n — the numbers used in formula.
Constraints
2 ≤ x ≤ 1000000000
1 ≤ n ≤ 1000000000000000000Print the answer corresponding to the input.Sample Input 1
10 2
Sample Output 1
2
Sample Input 2
20190929 1605
Sample Output 2
363165664
Explanation
In the first example, g(10,1)=f(1,2)⋅f(1,5)=1, g(10,2)=f(2,2)⋅f(2,5)=2., I have written this Solution Code: import math
mod = 1000000007
def modExpo(x, n):
if n <= 0:
return 1
if n % 2 == 0:
return modExpo((x * x) % mod, n // 2)
return (x * modExpo((x * x) % mod, (n - 1) // 2)) % mod
def calc(n, p):
prod = 1
pPow = 1
while pPow <= (n // p):
pPow *= p
add = 0
while pPow != 1:
quo = n // pPow
quo -= add
power = modExpo(pPow, quo)
prod = (prod * power) % mod
add += quo
pPow //= p
return prod
x, n = map(int, input().split())
res = 1
i = 2
while (i * i) <= x:
if x % i == 0:
res = (res * calc(n, i)) % mod
while x % i == 0:
x //= i
i += 1
if x > 1:
res = (res * calc(n, x)) % mod
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's assume some functional definitions for this problem.
We take prime(x) as the set of all prime divisors of x. For example, prime(140)={2,5,7}, prime(169)={13}.
Let f(x,p) be the maximum possible integer p**k where k is an integer such that x is divisible by p**k.
(Here a**b means a raised to the power b or pow(a, b))
For example:
f(99,3)=9 (99 is divisible by 3**2=9 but not divisible by 3**3=27),
f(63,7)=7 (63 is divisible by 7**1=7 but not divisible by 7**2=49).
Let g(x,y) be the product of f(y,p) for all p in prime(x).
For example:
g(30,70)=f(70,2)*f(70,3)*f(70,5)=2*1*5=10,
g(525,63)=f(63,3)*f(63,5)*f(63,7)=9*1*7=63.
You are given two integers x and n. Calculate g(x,1)*g(x,2)*…*g(x,n) mod (1000000007).
(Read modulo exponentiation before attempting this problem)The only line contains integers x and n — the numbers used in formula.
Constraints
2 ≤ x ≤ 1000000000
1 ≤ n ≤ 1000000000000000000Print the answer corresponding to the input.Sample Input 1
10 2
Sample Output 1
2
Sample Input 2
20190929 1605
Sample Output 2
363165664
Explanation
In the first example, g(10,1)=f(1,2)⋅f(1,5)=1, g(10,2)=f(2,2)⋅f(2,5)=2., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int power(int a, int b){
int ans = 1;
b %= (mod-1);
while(b){
if(b&1)
ans = (ans*a) % mod;
b >>= 1;
a = (a*a) % mod;
}
return ans;
}
int f(int n, int p){
int ans = 1;
int cur = 1;
while(cur <= n/p){
cur = cur*p;
int z = power(p, n/cur);
ans = (ans*z) % mod;
}
return ans;
}
signed main() {
IOS;
int x, n, ans = 1;
cin >> x >> n;
for(int i = 2; i*i <= x; i++){
if(x%i != 0) continue;
ans = (ans*f(n, i)) % mod;
while(x%i == 0)
x /= i;
}
if(x > 1)
ans = (ans*f(n, x)) % mod;
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |