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For this Question: Given an array of integers, and an integer ‘K’, find the count of pairs of elements in the array whose sum is equal to 'K'.The first line of the input contains 2 space separated integers N and K denoting the size of array and the sum respectively.
The second line of the input contains N space separated integers denoting the elements of the array.
Constraints:
1 <= N <= 100000
1 <= K <= 100000
1 <= A[i] <= 100000Print the count of pairs of elements in the array whose sum is equal to the K.Sample Input:
4 6
1 5 7 1
Sample Output:
2, I have written this Solution Code: n,k=input().split()
n=int(n)
k=int(k)
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
x=[0]*100000
for i in range(0,n):
x[arr[i]]+=1
count=0
for i in range(0,n):
count+=x[k-arr[i]]
if((k-arr[i])==arr[i]):
count-=1
print (int(count/2)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers, and an integer ‘K’, find the count of pairs of elements in the array whose sum is equal to 'K'.The first line of the input contains 2 space separated integers N and K denoting the size of array and the sum respectively.
The second line of the input contains N space separated integers denoting the elements of the array.
Constraints:
1 <= N <= 100000
1 <= K <= 100000
1 <= A[i] <= 100000Print the count of pairs of elements in the array whose sum is equal to the K.Sample Input:
4 6
1 5 7 1
Sample Output:
2, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define read(type) readInt<type>()
#define max1 1000008
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ull cnt[max1];
int main(){
int n,k;
cin>>n>>k;
int a[n];
unordered_map<int,int> m;
ll ans=0;
for(int i=0;i<n;i++){
cin>>a[i];
if(m.find(k-a[i])!=m.end()){ans+=m[k-a[i]];}
m[a[i]]++;
}
cout<<ans<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers, and an integer ‘K’, find the count of pairs of elements in the array whose sum is equal to 'K'.The first line of the input contains 2 space separated integers N and K denoting the size of array and the sum respectively.
The second line of the input contains N space separated integers denoting the elements of the array.
Constraints:
1 <= N <= 100000
1 <= K <= 100000
1 <= A[i] <= 100000Print the count of pairs of elements in the array whose sum is equal to the K.Sample Input:
4 6
1 5 7 1
Sample Output:
2, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int arrSize = sc.nextInt();
int targetK = sc.nextInt();
int arr[] = new int[arrSize];
for(int i = 0; i < arrSize; i++)
arr[i] = sc.nextInt();
System.out.println(countPairs(arr, arrSize, targetK));
}
static long countPairs(int arr[], int arrSize, int targetK)
{
long ans = 0;
HashMap<Integer, Integer> hash = new HashMap<>();
for(int i = 0; i < arrSize; i++)
{
int elem = arr[i];
if(hash.containsKey(targetK-elem) == true)
ans += hash.get(targetK-elem);
if(hash.containsKey(elem) == true)
{
int freq = hash.get(elem);
hash.put(elem, freq+1);
}
else
hash.put(elem, 1);
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., 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>
#define rep(i,n) for (int i=0; i<(n); i++)
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
int ans=0;
int v=0;
for(int i=1;i<n;++i){
if(a[i]>a[i-1])
v=0;
else
++v;
ans=max(ans,v);
}
cout<<ans;
#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: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., 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 num = Integer.parseInt(br.readLine());
String s= br.readLine();
int[] arr= new int[num];
String[] s1 = s.split(" ");
int ccount = 0, pcount = 0;
for(int i=0;i<(num);i++)
{
arr[i]=Integer.parseInt(s1[i]);
if(i+1 < num){
arr[i+1] = Integer.parseInt(s1[i+1]);}
if(((i+1)< num) && (arr[i]>=arr[i+1]) ){
ccount++;
}else{
if(ccount > pcount){
pcount = ccount;
}
ccount = 0;
}
}
System.out.print(pcount);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., I have written this Solution Code: N = int(input())
arr = iter(map(int,input().strip().split()))
jumps = []
jump = 0
temp = next(arr)
for i in arr:
if i<=temp:
jump += 1
temp = i
continue
temp = i
jumps.append(jump)
jump = 0
jumps.append(jump)
print(max(jumps)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Asta wants to become a magic knight. In his journey to becoming a magic knight, he stuck on a problem and asks for your help. Problem description:-
Given a number N which at the end of each second gets halved(take floor value) and then increases by 2. Your task is to calculate the number N at the end of T second.<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>MagicKnight()</b> that takes integers N and T as parameters.
Constraints:-
3 <= N <= 1000000000
1 <= T <= 1000000000Return the number N at the end of T seconds.Sample Input:-
7 2
Sample Output:-
3
Explanation:-
After 1 sec :- N = 7/2 + 2 = 5
After 2 sec :- N = 5/2 + 2 = 4
Sample Input:-
10 1
Sample Output:-
7, I have written this Solution Code: static int MagicKnight(int N, int T){
while(T-->0){
N/=2;
N+=2;
if(N==3 || N==4){return N;}
}
return N;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Asta wants to become a magic knight. In his journey to becoming a magic knight, he stuck on a problem and asks for your help. Problem description:-
Given a number N which at the end of each second gets halved(take floor value) and then increases by 2. Your task is to calculate the number N at the end of T second.<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>MagicKnight()</b> that takes integers N and T as parameters.
Constraints:-
3 <= N <= 1000000000
1 <= T <= 1000000000Return the number N at the end of T seconds.Sample Input:-
7 2
Sample Output:-
3
Explanation:-
After 1 sec :- N = 7/2 + 2 = 5
After 2 sec :- N = 5/2 + 2 = 4
Sample Input:-
10 1
Sample Output:-
7, I have written this Solution Code: long MagicKnight(long N, long T){
while(T--){
N/=2;
N+=2;
if(N==3 || N==4){return N;}
}
return N;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Asta wants to become a magic knight. In his journey to becoming a magic knight, he stuck on a problem and asks for your help. Problem description:-
Given a number N which at the end of each second gets halved(take floor value) and then increases by 2. Your task is to calculate the number N at the end of T second.<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>MagicKnight()</b> that takes integers N and T as parameters.
Constraints:-
3 <= N <= 1000000000
1 <= T <= 1000000000Return the number N at the end of T seconds.Sample Input:-
7 2
Sample Output:-
3
Explanation:-
After 1 sec :- N = 7/2 + 2 = 5
After 2 sec :- N = 5/2 + 2 = 4
Sample Input:-
10 1
Sample Output:-
7, I have written this Solution Code: def MagicKnight(N,T):
while T > 0:
N = N//2
N = N+2
if N == 3 or N ==4:
return N
T = T - 1
return N
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Asta wants to become a magic knight. In his journey to becoming a magic knight, he stuck on a problem and asks for your help. Problem description:-
Given a number N which at the end of each second gets halved(take floor value) and then increases by 2. Your task is to calculate the number N at the end of T second.<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>MagicKnight()</b> that takes integers N and T as parameters.
Constraints:-
3 <= N <= 1000000000
1 <= T <= 1000000000Return the number N at the end of T seconds.Sample Input:-
7 2
Sample Output:-
3
Explanation:-
After 1 sec :- N = 7/2 + 2 = 5
After 2 sec :- N = 5/2 + 2 = 4
Sample Input:-
10 1
Sample Output:-
7, I have written this Solution Code: long MagicKnight(long N, long T){
while(T--){
N/=2;
N+=2;
if(N==3 || N==4){return N;}
}
return N;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The underage Harry and Cedric reach the Triwizard Cup together in the hedge maze, which turns out to be a portkey that transports them to a graveyard where Pettigrew and Voldemort are waiting. After Pettigrew murders Cedric with the Killing Curse on Voldemort's order, Voldemort challenges Harry with a very tricky spell, <b>The string leviosa</b>.
In order to survive this spell, Harry has to answer the following question:
There is a string of size N consisting of characters A, B, and C. The character A can be transformed to either character B or character C. Now, two players play a game with the following rules turn wise. In each turn, the player with the turn chooses a position with character A in the string and transforms the character to another character (B or C). The only restriction to the string is that there can't be two consecutive <b>B</b>s or two consecutive <b>C</b>s in the string. The player who cannot make a move loses. Which player wins the game?
Can you help Harry in surviving the dangerous spell?The first line of the input contains a single integer N, the length of the string.
The second line of the input contains N characters of the string.
Constraints
1 <= N <= 100000
The characters in the string are restricted to A, B, and C.
There are no consecutive Bs or Cs in the input string.Output "Player1" if the player who starts the game wins, else output "Player2" (without quotes).Sample Input
3
ACA
Sample Output
Player2
Explanation: Player1 changes the first or the last character to B, the second player changes the remaining A to B in the next turn. After that, player1 cannot make a move since there are no more As in the string.
Sample Input
4
BACA
Sample Output
Player1
Explanation: Player1 changes the last character in the string to B and player2 cannot make a move further., I have written this Solution Code: n = int(input())
input_string = list(input())
win = False
if len(input_string) == 1:
if input_string[0] == "A":
win = not win
else:
for i in range(n):
if input_string[i] == "A":
if i == 0:
if input_string[i + 1] == "B":
input_string[i] = "C"
win = not win
elif input_string[i + 1] == "C":
input_string[i] = "B"
win = not win
elif i == n - 1:
if input_string[i - 1] == "B":
input_string[i] = "C"
win = not win
elif input_string[i - 1] == "C":
input_string[i] = "B"
win = not win
else:
if input_string[i - 1] == input_string[i + 1]:
if input_string[i - 1] == "B":
input_string[i] = "C"
win = not win
elif input_string[i - 1] == "C":
input_string[i] = "B"
win = not win
else:
input_string[i] = "B"
win = not win
else:
if (input_string[i - 1] == "B" and input_string[i + 1] == "C") or (input_string[i - 1] == "C" and input_string[i + 1] == "B"):
input_string[i] = "X"
elif input_string[i - 1] == "B" and input_string[i + 1] == "A":
input_string[i] = "C"
win = not win
elif input_string[i - 1] == "C" and input_string[i + 1] == "A":
input_string[i] = "B"
win = not win
if win:
print("Player1")
else:
print("Player2"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The underage Harry and Cedric reach the Triwizard Cup together in the hedge maze, which turns out to be a portkey that transports them to a graveyard where Pettigrew and Voldemort are waiting. After Pettigrew murders Cedric with the Killing Curse on Voldemort's order, Voldemort challenges Harry with a very tricky spell, <b>The string leviosa</b>.
In order to survive this spell, Harry has to answer the following question:
There is a string of size N consisting of characters A, B, and C. The character A can be transformed to either character B or character C. Now, two players play a game with the following rules turn wise. In each turn, the player with the turn chooses a position with character A in the string and transforms the character to another character (B or C). The only restriction to the string is that there can't be two consecutive <b>B</b>s or two consecutive <b>C</b>s in the string. The player who cannot make a move loses. Which player wins the game?
Can you help Harry in surviving the dangerous spell?The first line of the input contains a single integer N, the length of the string.
The second line of the input contains N characters of the string.
Constraints
1 <= N <= 100000
The characters in the string are restricted to A, B, and C.
There are no consecutive Bs or Cs in the input string.Output "Player1" if the player who starts the game wins, else output "Player2" (without quotes).Sample Input
3
ACA
Sample Output
Player2
Explanation: Player1 changes the first or the last character to B, the second player changes the remaining A to B in the next turn. After that, player1 cannot make a move since there are no more As in the string.
Sample Input
4
BACA
Sample Output
Player1
Explanation: Player1 changes the last character in the string to B and player2 cannot make a move further., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n;
string s;
cin >> n >> s;
assert(sz(s)==n);
int l = 0, r = 1;
int ans = 0;
while (l < n) {
while (l < n && s[l] != 'A') ++l, ++r;
while (r < n && s[r] == 'A') ++r;
if (l == 0 && r == n) {
ans ^= n % 2;
}
else if (r == n || l == 0) {
ans ^= r - l;
}
else if (r < n && s[l - 1] == s[r]){
ans ^= 1;
}
l = r;
}
cout << (ans ? "Player1" : "Player2");
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Is it possible to get a sum of <b>B</b> when throwing a die with six faces 1, 2, …, 6 <b>A</b> times?The input consists of two space-separated integers.
A B
<b>Constraints</b>
1≤A≤100
1≤B≤1000
A and B are integers.If it is possible to get a sum of B, print Yes; otherwise, print No.<b>Sample Input 1</b>
2 11
<b>Sample Output 1</b>
Yes
<b>Sample Input 2</b>
2 13
<b>Sample Output 2</b>
No, I have written this Solution Code: #include <iostream>
using namespace std;
int main() {
int A, B;
cin >> A >> B;
if(A <= B && B <= 6 * A) {
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a side of a square, your task is to calculate and print its area.The first line of the input contains the side of the square.
<b>Constraints:</b>
1 <= side <=100You just have to print the area of a squareSample Input:-
3
Sample Output:-
9
Sample Input:-
6
Sample Output:-
36, I have written this Solution Code: def area(side_of_square):
print(side_of_square*side_of_square)
def main():
N = int(input())
area(N)
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a side of a square, your task is to calculate and print its area.The first line of the input contains the side of the square.
<b>Constraints:</b>
1 <= side <=100You just have to print the area of a squareSample Input:-
3
Sample Output:-
9
Sample Input:-
6
Sample Output:-
36, 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 side = Integer.parseInt(br.readLine());
System.out.print(side*side);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: To seek revenge on Midgard, Loki devises a plan to torture the humans by making them take part in one of his silly games. He makes N people sit in a circle. He says he will kill every kth person sitting in the circle, starting from 1st person.
Loki performs his revenge prank until and unless 1 survivor remains.
What is the initial position of the survivor, if the indexing is clockwise?The first line of input contains a single integer T. The next T line of input contains Two space separated integers each containing value of N and k.
<b>Constraints:</b>
1 <= T <= 100
1 <= k, N <= 20Print the initial position of the survivor.Sample Input:
2
3 2
5 3
Sample Output
3
4
Explanation:
Test case 1: There are 3 people so skipping 1 person i.e 1st person 2nd person will be killed in next step 3rd person will be skipped and 1st person will be killed. Thus the safe position is 3.
Test case 2: 2 people i.e 1and 2 are skipped and person 3 will be killed in next step 4 and 5 will be skipped and 1st person will be killed next step 2 and 4 will be skipped and 5th person will be killed next step first 2 will be skipped then 4 will be skipped and so coming back to 2 therefore person 2 will be killed. Thus the safe position is 4., I have written this Solution Code: t=int(input())
while t>0:
n,k=map(int,input().split())
s=set()
remove=0
for i in range(1,n+1):
s.add(i)
i=1
while True:
if len(s)==1:
break
if i in s:
remove+=1
if remove==k:
if i in s:
s.remove(i)
remove=0
i+=1
if i>n:
i=1
print(*s)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: To seek revenge on Midgard, Loki devises a plan to torture the humans by making them take part in one of his silly games. He makes N people sit in a circle. He says he will kill every kth person sitting in the circle, starting from 1st person.
Loki performs his revenge prank until and unless 1 survivor remains.
What is the initial position of the survivor, if the indexing is clockwise?The first line of input contains a single integer T. The next T line of input contains Two space separated integers each containing value of N and k.
<b>Constraints:</b>
1 <= T <= 100
1 <= k, N <= 20Print the initial position of the survivor.Sample Input:
2
3 2
5 3
Sample Output
3
4
Explanation:
Test case 1: There are 3 people so skipping 1 person i.e 1st person 2nd person will be killed in next step 3rd person will be skipped and 1st person will be killed. Thus the safe position is 3.
Test case 2: 2 people i.e 1and 2 are skipped and person 3 will be killed in next step 4 and 5 will be skipped and 1st person will be killed next step 2 and 4 will be skipped and 5th person will be killed next step first 2 will be skipped then 4 will be skipped and so coming back to 2 therefore person 2 will be killed. Thus the safe position is 4., I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.*;
class Main{
public static void main (String[] args)throws IOException
{
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while(t-->0)
{
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
System.out.println(safe_Position(n,k));
}
}
public static int safe_Position(int n, int k)
{
if (n == 1) //base case
return 1;
else
/* The position returned by safe_Position(n - 1, k) is adjusted because the
recursive call safe_Position(n - 1, k) considers the original position
k%n + 1 as position 1 */
return (safe_Position(n - 1, k) + k-1) % n + 1; //recursion
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Check whether the number N is representable as the sum of the cubes of two positive integers i. e. check whether there exist two positive integers a and b such that a^3 + b^3 = N.First line contains an integer N.
Constraints
1 <= N <= 10^12Print "YES" if N is representable as the sum of the cubes of two positive integers otherwise "NO".Sample Input 1:
3
Output
NO
Explanation:
3 cannot be represented as sum of two cubes.
Sample Input 2:
35
Output
YES
Explanation:
35 = 8 + 27 = 2^3 + 3^3 , I have written this Solution Code: import math
def solve(n):
lo = 1
hi = round(math.pow(n, 1 / 3))
while (lo <= hi):
curr = (lo * lo * lo + hi * hi * hi)
if (curr == n):
return True
if (curr < n):
lo += 1
else:
hi -= 1
return False
n=int(input())
if solve(n):
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 positive integer N. Check whether the number N is representable as the sum of the cubes of two positive integers i. e. check whether there exist two positive integers a and b such that a^3 + b^3 = N.First line contains an integer N.
Constraints
1 <= N <= 10^12Print "YES" if N is representable as the sum of the cubes of two positive integers otherwise "NO".Sample Input 1:
3
Output
NO
Explanation:
3 cannot be represented as sum of two cubes.
Sample Input 2:
35
Output
YES
Explanation:
35 = 8 + 27 = 2^3 + 3^3 , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
static boolean binary_search(long a[], int n, long tar) {
int l=0;
int r = n - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (a[mid] == tar) {
return true;
}
if (a[mid] < tar)l = mid + 1;
else r = mid - 1;
}
return false;
}
public static void main (String args[]) throws IOException
{
BufferedReader br = new BufferedReader (new InputStreamReader(System.in));
long n=Long.parseLong(br.readLine());
long a[] = new long[10000];
for (int i = 1; i <= 10000; i++) {
a[i-1] = (long)i*(long)i*(long)i;
}
boolean ans = false;
for (long i = 1; i <= 10000; i++) {
if (binary_search(a, 10000, n - i * i * i)) {
ans = true;
break;
}
}
if (ans) {
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: Given an unsorted array A[] of integers and size N, and an element X, you need to find the Minimum Number of Elements required to be added to this array so the new median of array becomes X.
Median of array is middle element of array in its sorted form. For odd number of elements N it is the element at position (N-1)/2. For even number of elements N, it is (A[N/2] + A[(N/2) - 1])/2.
Try without sorting the arrayThe first line contains number of testcases T. Every testcase consists of 2 lines, first line contains N - no of element of arrays and X, and second line contains Array elements in unsorted manner.
Constraints:
1 <= T <= 100
1 <= N <= 2*10^5
1 <= X <= 10^6
1 <= A[i] <= 10^6
Sum of N for every test case is less than or equal to 2*10^5You need to find total number of elements to be added to array so that the new median becomes equal to X.Sample Input:
2
6 30
10 20 30 100 150 200
5 50
10 20 30 100 150
Sample Output:
1
1
Explanation:
Testcase 1: Only 1 element before 30 is required to be added to the array, to make median of array 30.
Testcase 2: Only 1 element between 30 and 100, i. e number 70 is required to be added to make median 50., I have written this Solution Code: for i in range(int(input())):
n,k=map(int,input().split())
x=list(map(int,input().split()))
count,large_count,small_count,small,large=0,0,0,-1,10000000
for i in x:
if i==k:
count+=1
if i>k:
large_count+=1
large=min(large,i)
if i<k:
small_count+=1
small=max(small,i)
ans=abs(large_count-small_count)
if(count>0):
if(count>=n/2):
print(0)
else:
print(ans)
elif(small_count==large_count):
if((large+small)//2==k):
print(0)
else:
print(1)
elif(small_count==0 or large_count==0):
print(n)
else:
t=2*k-small
if(t==large):
print(ans)
elif(t<large):
large_count+=1
print(min(ans+1,abs(large_count-small_count)+1))
elif(t>large):
l=2*k-large
if(l>small):
small_count+=1
print(min(ans+1,abs(large_count-small_count)+1))
else:
print(ans+1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A[] of integers and size N, and an element X, you need to find the Minimum Number of Elements required to be added to this array so the new median of array becomes X.
Median of array is middle element of array in its sorted form. For odd number of elements N it is the element at position (N-1)/2. For even number of elements N, it is (A[N/2] + A[(N/2) - 1])/2.
Try without sorting the arrayThe first line contains number of testcases T. Every testcase consists of 2 lines, first line contains N - no of element of arrays and X, and second line contains Array elements in unsorted manner.
Constraints:
1 <= T <= 100
1 <= N <= 2*10^5
1 <= X <= 10^6
1 <= A[i] <= 10^6
Sum of N for every test case is less than or equal to 2*10^5You need to find total number of elements to be added to array so that the new median becomes equal to X.Sample Input:
2
6 30
10 20 30 100 150 200
5 50
10 20 30 100 150
Sample Output:
1
1
Explanation:
Testcase 1: Only 1 element before 30 is required to be added to the array, to make median of array 30.
Testcase 2: Only 1 element between 30 and 100, i. e number 70 is required to be added to make median 50., 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;
int a = 0, b = 0, c = 0;
int l = 0, r = inf;
for(int i = 1; i <= n; i++){
int p; cin >> p;
if(p < x) a++, l = max(l, p);
else if(p == x) b++;
else c++, r = min(r, p);
}
//cout << l << " " << r << endl;
//cout << a << " " << b << " " << c << endl;
int ans;
if(n&1){
if(a <= n/2){
if(a+b > n/2)
ans = 0;
else{
if(2*x - r >= l && b == 0)
ans = 2*c - n;
else
ans = 2*c + 1 - n;
}
}
else{
if(2*x - l <= r && b == 0)
ans = 2*a - n;
else
ans = 2*a + 1 - n;
}
}
else{
if(a < n/2){
if(c == n)
ans = n;
else if(a+b > n/2)
ans = 0;
else{
if(2*x - r >= l && b == 0)
ans = 2*c - n;
else
ans = 2*c + 1 - n;
}
}
else if(a > n/2){
if(2*x - l <= r && b == 0)
ans = 2*a - n;
else
ans = 2*a + 1 - n;
}
else
ans = 1 - ((l+r) == 2*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: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 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 a[N];
bool check_increasing(int n){
int id = 0;
for(int i = 2; i <= n; i++){
if(a[i] < a[i-1]){
id = i;
break;
}
}
if(id == 0) return 0;
int cur = id+1;
while(cur != id+n){
if(a[cur] <= a[cur-1])
return 0;
cur++;
}
return 1;
}
bool check_decreasing(int n){
int id = 0;
for(int i = 2; i <= n; i++){
if(a[i] > a[i-1]){
id = i;
break;
}
}
if(id == 0) return 0;
int cur = id+1;
while(cur != id+n){
if(a[cur] >= a[cur-1])
return 0;
cur++;
}
return 1;
}
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i], a[i+n] = a[i];
if(check_increasing(n))
cout << "Yes" << endl;
else if(check_decreasing(n))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code:
cases = int(input())
for _ in range(cases):
size = int(input())
orig = list(map(int,input().split()))
givenList = orig.copy()
givenList.sort()
flag = False
for _ in range(size-1):
ele = givenList.pop()
givenList.insert(0,ele)
if givenList == orig:
print("Yes")
flag = True
break
if not flag:
givenList.sort()
givenList.reverse()
for _ in range(size-1):
ele = givenList.pop()
givenList.insert(0,ele)
if givenList == orig:
print("Yes")
flag = True
break
if not flag:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim()); //Inputting the testcases
while(t-->0){
long n = Long.parseLong(br.readLine());
int arr[] = new int[(int)n];
String inputLine[] = br.readLine().trim().split("\\s+");
for(long i=0; i<n; i++){
arr[(int)i] = Integer.parseInt(inputLine[(int)i]);
}
long mini = Integer.MAX_VALUE, maxi = Integer.MIN_VALUE;
long max_index = 0, min_index = 0;
for(long i=0; i<n; i++){
if(maxi < arr[(int)i]){
maxi = arr[(int)i];
max_index = i;
}
if(mini > arr[(int)i]){
mini = arr[(int)i];
min_index = i;
}
}
int flag = 0;
if(max_index == min_index -1)
flag = 1;
else if(min_index == max_index - 1)
flag = -1;
if(flag == 1){
for(long i = 1; flag==1 && i<=max_index; ++i){
if(arr[(int)i-1] >= arr[(int)i])
flag = 0;
}
for(long i = min_index+1; flag==1 && i<n; ++i){
if(arr[(int)i-1] >= arr[(int)i])
flag = 0;
}
if(arr[0]<=arr[(int)n-1])
flag = 0;
} else if(flag == -1){
for(long i = 1; flag ==-1 && i<=min_index; ++i){
if(arr[(int)i-1] <= arr[(int)i])
flag = 0;
}
for(long i = max_index+1; flag==-1 && i<n; ++i){
if(arr[(int)i-1] <= arr[(int)i])
flag = 0;
}
if(arr[0]>=arr[(int)n-1])
flag = 0;
}
if(flag == 0)
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 natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<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>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<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>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: static int focal_length(int R, char Mirror)
{
int f=R/2;
if((R%2==1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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 integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: def Pattern(N):
print('*')
for i in range (0,N-2):
print('*',end='')
for j in range (0,i+1):
print('^',end='')
print('*')
for i in range (0,N+1):
print('*',end='')
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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 integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
cout<<'*'<<endl;
for(int i=0;i<N-2;i++){
cout<<'*';
for(int j=0;j<=i;j++){
cout<<'^';
}
cout<<'*'<<endl;
}
for(int i=0;i<=N;i++){
cout<<'*';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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 integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: static void Pattern(int N){
System.out.println('*');
for(int i=0;i<N-2;i++){
System.out.print('*');
for(int j=0;j<=i;j++){
System.out.print('^');
}System.out.println('*');
}
for(int i=0;i<=N;i++){
System.out.print('*');
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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 integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1073741824
#define read(type) readInt<type>()
#define max1 1000001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's little
// theorem.
ll ncr(ll n, ll r,ll p)
{
// Base case
if (r==0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
ll fac[n+1];
fac[0] = 1;
for (ll i=1 ; i<=n; i++)
fac[i] = fac[i-1]*i%p;
return (fac[n]* modInverse(fac[r], p) % p *
modInverse(fac[n-r], p) % p) % p;
}
ll fastexp (ll a, ll b, ll n) {
ll res = 1;
while (b) {
if (b & 1) res = res*a%n;
a = a*a%n;
b >>= 1;
}
return res;
}
bool a[max1];
int main() {
FOR(i,max1){
a[i]=false;}
for(int i=2;i<max1;i++){
if(a[i]==false){
for(int j=i+i;j<max1;j+=i){
a[j]=true;
}
}}
vector<int> v;
map<int,int> m;
for(int i=2;i<max1;i++){
if(a[i]==false){
v.EB(i);
m[i]++;
}
}
int t;
cin>>t;
while(t--){
int n;
cin>>n;
bool win=false;
for(int i=0;i<v.size();i++){
if(m.find(n-v[i])!=m.end()){
if(v[i]>n){break;}
cout<<v[i]<<" "<<n-v[i]<<endl;win=true;break;
}
}
if(win==false){out(-1);}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static boolean isPrime(int n)
{
boolean ans=true;
if(n<=1)
{
ans=false;
}
else if(n==2 || n==3)
{
ans=true;
}
else if(n%2==0 || n%3==0)
{
ans=false;
}
else
{
for(int i=5;i*i<=n;i+=6)
{
if(n%i==0 || n%(i+2)==0)
{
ans=false;
break;
}
}
}
return ans;
}
public static void main (String[] args) throws IOException {
BufferedReader scan=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(scan.readLine());
while(t-->0)
{
int n=Integer.parseInt(scan.readLine());
int a=0,b=0;
if(n%2==1)
{
if(isPrime(n-2))
{
a=2;b=n-2;
}
}
else
{
for(int i=2;i<=n/2;i++)
{
if(isPrime(i))
{
if(isPrime(n-i))
{
a=i;b=n-i;
break;
}
}
}
}
if(a!=0 && b!=0)
{
System.out.println(a+" "+b);
}
else
{
System.out.println(-1);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.
NOTE: A solution will always exist if the number is even. Read Goldbach’s conjecture.
If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.
Constraints:
1 ≤ T ≤ 100
2 ≤ N ≤ 1000000Print the two prime numbers in a single line with space in between if exist else print -1.Sample Input:
2
8
3
Sample Output:
3 5
-1, I have written this Solution Code: def isprime(num):
if(num==1 or num==0):
return False
for i in range(2,int(num**0.5)+1):
if(num%i==0):
return False
return True
T = int(input())
for test in range(T):
N = int(input())
value = -1
if(N%2==0):
for i in range(2,int(N/2)+1):
if(isprime(i)):
if(isprime(N-i)):
value = i
break
else:
if(isprime(N-2)):
value = 2
if(value==-1):
print(value)
else:
print(str(value)+" "+str(N-value)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton loves playing the ancient game of Chips. It is played on a directed graph consisting of N vertices and M edges. Each vertex of the graph initially contains A<sub>i</sub> chips. A move is defined as:
Choose a vertex i, such that the number of chips at vertex i is exactly equal to the out-degree of vertex i. Also, the number of chips at vertex i must be positive. Then, for each edge from i to v, shift exactly one chip from vertex i to vertex v.
It is guaranteed that the graph does not contain directed cycles. A player's score is the number of moves that they perform. It can be shown that the score will always be finite. Print the maximum score that can be attained by a player.
Further, suppose the player is allowed to place extra chips on some of the vertices. Formally, if the player is allowed to increase the initial number of chips at each vertex i to B<sub>i</sub> (where B<sub>i</sub> ≥ A<sub>i</sub>), then print the minimum number of extra chips needed to increase the maximum score of the game. If it is not possible to increase the score of the game, print -1.The first line of the input contains the number of test cases, T (1 ≤ T ≤ 10<sup>5</sup>).
The first line of each test case consists of two integers N (1≤ N ≤ 10<sup>5</sup>) and M (0 ≤ M ≤ min(2×10<sup>5</sup>, N(N - 1)/2)) — the number of vertices and edges respectively.
The second line consists of N integers — A<sub>1</sub>, A<sub>2</sub>, ... A<sub>N</sub> (0 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>).
Then M lines follow, the i<sup>th</sup> line containing two integers u<sub>i</sub> and v<sub>i</sub> — denoting there is an edge from u<sub>i</sub> to v<sub>i</sub>. There are no self-loops or multiple edges in the given graph.
It is guaranteed that the sum of N over all test cases is less than 3×10<sup>5</sup> and the sum of M over all test cases is less than 4×10<sup>5</sup>. For each test case, print two space separated integers. The first integer denotes the maximum score of the game. The second integer denotes the minimum number of chips that must be added to some vertices so that the maximum score of the game increases. If the score of the game cannot be increased, print -1 instead.Sample Input 1:
2
2 1
1 1
1 2
2 1
2 2
1 2
Sample Output 1:
1 -1
0 -1
Sample Input 2:
1
4 3
1 0 0 2
1 2
3 2
2 4
Sample Output 2:
2 1, 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 = 6e5 + 10;
int n, m;
vvi adj, adjrev;
vi vis, topo;
void dfs (int cur)
{
vis[cur] = true;
for (int i: adj[cur])
if (!vis[i])
dfs(i);
topo.pb(cur);
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
read(t);
while(t--)
{
cin >> n >> m;
adj.assign(n + 1, vi(0));
adjrev.assign(n + 1, vi(0));
vis.assign(n + 1, 0);
readarr(a, n);
int d[n + 1] = {};
FOR (i, 1, m)
{
readb(x, y);
adj[x].pb(y);
adjrev[y].pb(x);
d[x]++;
}
topo.clear();
FOR (i, 1, n)
if (!vis[i])
dfs(i);
reverse(all(topo));
int dp[n + 1] = {}, ans = 0, mn = inf;
for (int i: topo)
{
if (a[i] > d[i] || d[i] == 0) continue;
int sum = a[i];
for (int j: adjrev[i]) sum += dp[j];
dp[i] = sum/d[i];
ans += dp[i];
int x = d[i] - sum%d[i];
if (x + a[i] <= d[i]) mn = min(mn, x);
}
cout << ans << " ";
if (mn < inf) cout << mn << endl;
else cout << -1 << endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: a, b, v = map(int, input().strip().split(" "))
c = abs(a-b)
t = c//v
print(t), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int k = sc.nextInt();
System.out.print(Time(n,m,k));
}
static int Time(int A, int B, int S){
if(B>A){
return (B-A)/S;
}
return (A-B)/S;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
#define ma 10000001
bool a[ma];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++){
a[i]=false;
}
for(int i=2;i<=n;i++){
if(a[i]==false){
for(int j=i+i;j<=n;j+=i){
a[j]=true;
}
}
}
int cnt=0;
for(int i=2;i<=n;i++){
if(a[i]==false){cnt++;}
}
cout<<cnt;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n = Integer.parseInt(br.readLine());
long i=2,j,count,noOfPrime=0;
if(n<=1)
System.out.println("0");
else{
while(i<=n)
{
count=0;
for(j=2; j<=Math.sqrt(i); j++)
{
if( i%j == 0 ){
count++;
break;
}
}
if(count==0){
noOfPrime++;
}
i++;
}
System.out.println(noOfPrime);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: function numberOfPrimes(N)
{
let arr = new Array(N+1);
for(let i = 0; i <= N; i++)
arr[i] = 0;
for(let i=2; i<= N/2; i++)
{
if(arr[i] === -1)
{
continue;
}
let p = i;
for(let j=2; p*j<= N; j++)
{
arr[p*j] = -1;
}
}
//console.log(arr);
let count = 0;
for(let i=2; i<= N; i++)
{
if(arr[i] === 0)
{
count++;
}
}
//console.log(arr);
return count;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import math
n = int(input())
n=n+1
if n<3:
print(0)
else:
primes=[1]*(n//2)
for i in range(3,int(math.sqrt(n))+1,2):
if primes[i//2]:primes[i*i//2::i]=[0]*((n-i*i-1)//(2*i)+1)
print(sum(primes)), In this Programming Language: Python, 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: 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: Naruto and Sasuke are playing a game in which they are given a binary string S. The rules of the game are given as:-
Players play in turns
The game ends when all the characters in the string are same, and the player who has the current turn wins.
The current player has to delete the i<sup>th</sup> character from the string. (1 <= i <= |S|)
If both the players play optimally and Naruto goes first, find who will win the game.The Input contains only string S.
Constraints:-
1 <= |S| <= 100000Print "Naruto" if Naruto wins else print "Sasuke".Sample Input:-
0101
Sample Output:-
Sasuke
Explanation:-
First Naruto will delete the character 0 from index 1. string:- 101
Then Sasuke will delete the character 1 from index 1. string 01
It doesn't matter which character Naruto removes now, Sasuke will definitely win.
Sample Input:-
1111
Sample Output:-
Naruto
Explanation:-
All the characters in the string are already same hence Naruto wins, 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 s1=br.readLine();
int N=s1.length();
int index=0;
int count=0;
for(int i=N-2;i>=0;i--)
{
if(s1.charAt(i)!=s1.charAt(i+1))
{
index=i+1;
break;
}
}
if(index==N-1)
{
if(N%2==0)
{
System.out.print("Sasuke");
}
else{
System.out.print("Naruto");
}
}
else if(index==0)
{
System.out.print("Naruto");
}
else{
for(int i=0;i<index;i++)
{
count++;
}
if(count%2==0)
{
System.out.print("Naruto");
}
else{
System.out.print("Sasuke");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are playing a game in which they are given a binary string S. The rules of the game are given as:-
Players play in turns
The game ends when all the characters in the string are same, and the player who has the current turn wins.
The current player has to delete the i<sup>th</sup> character from the string. (1 <= i <= |S|)
If both the players play optimally and Naruto goes first, find who will win the game.The Input contains only string S.
Constraints:-
1 <= |S| <= 100000Print "Naruto" if Naruto wins else print "Sasuke".Sample Input:-
0101
Sample Output:-
Sasuke
Explanation:-
First Naruto will delete the character 0 from index 1. string:- 101
Then Sasuke will delete the character 1 from index 1. string 01
It doesn't matter which character Naruto removes now, Sasuke will definitely win.
Sample Input:-
1111
Sample Output:-
Naruto
Explanation:-
All the characters in the string are already same hence Naruto wins, I have written this Solution Code: binary_string = input()
n = len(binary_string)-1
index = n
for i in range(n-1, -1, -1):
if binary_string[i] == binary_string[n]:
index = i
else:
break
if index % 2 == 0:
print("Naruto")
else:
print("Sasuke"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are playing a game in which they are given a binary string S. The rules of the game are given as:-
Players play in turns
The game ends when all the characters in the string are same, and the player who has the current turn wins.
The current player has to delete the i<sup>th</sup> character from the string. (1 <= i <= |S|)
If both the players play optimally and Naruto goes first, find who will win the game.The Input contains only string S.
Constraints:-
1 <= |S| <= 100000Print "Naruto" if Naruto wins else print "Sasuke".Sample Input:-
0101
Sample Output:-
Sasuke
Explanation:-
First Naruto will delete the character 0 from index 1. string:- 101
Then Sasuke will delete the character 1 from index 1. string 01
It doesn't matter which character Naruto removes now, Sasuke will definitely win.
Sample Input:-
1111
Sample Output:-
Naruto
Explanation:-
All the characters in the string are already same hence Naruto wins, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int cnt[max1];
signed main(){
string s;
cin>>s;
int cnt=0;
FOR(i,s.length()){
if(s[i]=='0'){cnt++;}
}
if(cnt==s.length() || cnt==0){out("Naruto");return 0;}
if(s.length()%2==0){
out("Sasuke");
}
else{
out("Naruto");
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: static boolean isArmstrong(int N)
{
int num = N;
int sum = 0;
while(N > 0)
{
int digit = N%10;
sum += digit*digit*digit;
N = N/10;
}
if(num == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: function isArmstrong(n) {
// write code here
// do no console.log the answer
// return the output using return keyword
let sum = 0
let k = n;
while(k !== 0){
sum += Math.pow( k%10,3)
k = Math.floor(k/10)
}
return sum === n
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 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{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bob has a fantasy for integer 200. He has an integer N and wants you to do the given operation K times and print the result.
<b>Operation:</b>
1. Add 200 to the end of it, if N is not divisible by 200.
2. Otherwise divide it by 200.
For example, 6 would become 6200 and 434 would become 434200.The first line contains two integers N and K.
Constraints
All values in the input are integers.
1≤N≤10 ^ 5
1≤K≤20Print the answer as an integer.Sample Input 1
2021 4
Sample Output 1
50531
Applying the operation on N=2021 results in N becoming 2021→2021200→10106→10106200→50531.
Sample Input 2
40000 2
Sample Output 2
1
Sample Input 3
8691 20
Sample Output 3
84875488281
The answer may not fit into the signed 32- bit integer type., 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().split(" ");
long N=Long.parseLong(str[0]);
long K=Long.parseLong(str[1]);
for(int i=0; i<K; i++)
{
if(N % 200 == 0)
N /=200L;
else
N =(N *1000L) + 200L;
}
System.out.println(N);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bob has a fantasy for integer 200. He has an integer N and wants you to do the given operation K times and print the result.
<b>Operation:</b>
1. Add 200 to the end of it, if N is not divisible by 200.
2. Otherwise divide it by 200.
For example, 6 would become 6200 and 434 would become 434200.The first line contains two integers N and K.
Constraints
All values in the input are integers.
1≤N≤10 ^ 5
1≤K≤20Print the answer as an integer.Sample Input 1
2021 4
Sample Output 1
50531
Applying the operation on N=2021 results in N becoming 2021→2021200→10106→10106200→50531.
Sample Input 2
40000 2
Sample Output 2
1
Sample Input 3
8691 20
Sample Output 3
84875488281
The answer may not fit into the signed 32- bit integer type., I have written this Solution Code: a,b=input().split()
for i in range(int(b)):
a=str(a)+'200' if int(a)%200!=0 else int(a)//200
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bob has a fantasy for integer 200. He has an integer N and wants you to do the given operation K times and print the result.
<b>Operation:</b>
1. Add 200 to the end of it, if N is not divisible by 200.
2. Otherwise divide it by 200.
For example, 6 would become 6200 and 434 would become 434200.The first line contains two integers N and K.
Constraints
All values in the input are integers.
1≤N≤10 ^ 5
1≤K≤20Print the answer as an integer.Sample Input 1
2021 4
Sample Output 1
50531
Applying the operation on N=2021 results in N becoming 2021→2021200→10106→10106200→50531.
Sample Input 2
40000 2
Sample Output 2
1
Sample Input 3
8691 20
Sample Output 3
84875488281
The answer may not fit into the signed 32- bit integer type., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;++i)
const int inf = INT_MAX;
using ll = long long;
int main(){
long long int n,k;
cin >> n >> k;
rep(i,k){
if(n%200==0){
n/=200;
}else{
n=n*1000+200;
}
}
cout << n << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
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_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X find an integer Y such that Y has exactly two set bits in its binary representaion and abs(X- Y) is minimum.Input contains a single integer X.
Constraints
1 <= X <= 10^15Print a single integer, the minimum value of abs(X- Y).Sample input 1
5
Sample output 1
0
Explanation: Y=5.
Sample input 2
1
Sample output 2
2
Explanation: Y=3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
long N = Long.parseLong(br.readLine());
long minDiff = Long.MAX_VALUE;
for(long i = 0; i < 63; i++){
for(long j = (i+1); j < 63; j++){
long targetNumber = (1l<<i)|(1l<<j);
long diff = Math.abs(N - targetNumber);
if(diff < minDiff){
minDiff = diff;
}
}
}
System.out.println(minDiff);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X find an integer Y such that Y has exactly two set bits in its binary representaion and abs(X- Y) is minimum.Input contains a single integer X.
Constraints
1 <= X <= 10^15Print a single integer, the minimum value of abs(X- Y).Sample input 1
5
Sample output 1
0
Explanation: Y=5.
Sample input 2
1
Sample output 2
2
Explanation: Y=3, I have written this Solution Code: n = int(input())
ans = 10000000000000000
count = 0
temp = n
while (temp // 2 > 0):
count += 1
temp //= 2
if (n == 1 or n == 2 or n == 3):
print(3-n)
else:
for i in range(count + 1, -1, -1):
for j in range(i-1, -1, -1):
ans = min(ans, abs(n - ((1 << i) + (1 << j))))
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X find an integer Y such that Y has exactly two set bits in its binary representaion and abs(X- Y) is minimum.Input contains a single integer X.
Constraints
1 <= X <= 10^15Print a single integer, the minimum value of abs(X- Y).Sample input 1
5
Sample output 1
0
Explanation: Y=5.
Sample input 2
1
Sample output 2
2
Explanation: Y=3, 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>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int ans=1000000000000000;
for(int i=0;i<60;++i){
for(int j=i+1;j<60;++j){
ans=min(ans,abs(n-(1ll<<i)-(1ll<<j)));
}
}
cout<<ans;
#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: 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: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: N = int(input())
Nums = list(map(int,input().split()))
f = False
for n in Nums:
if n < 0:
f = True
break
if (f):
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 an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
bool win=false;
for(int i=0;i<n;i++){
cin>>a;
if(a<0){win=true;}}
if(win){
cout<<"Yes";
}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, 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 n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
for(int i=0;i<n;i++){
if(a[i]<0){System.out.print("Yes");return;}
}
System.out.print("No");
}
}
, 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: 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: Arun is known for his love of racing sports. He loves Formula 1 with a passion. He traveled to New Delhi to watch the Indian Grand Prix of this year. He noticed that one segment of the circuit was a long straight road. It was impossible for a car to overtake other cars on this segment. Therefore, a car had to lower down its speed if there was a slower car in front of it. While watching the race, Arun started to wonder how many cars were moving at their maximum speed.
Formally, you are provided the max speed of N cars in the sequence in which they entered the circuit's lengthy straightaway. Each car will prefer traveling at its max speed. If the front car's slowness prevents that from happening, it could have to slow down. Besides that, it continues to travel at its max speed while avoiding any crashes. You can assume that the straight section is infinitely long for the purposes of this problem. Count the number of vehicles that were traveling down the straight segment at their max speed.The first line of the input contains a single integer T denoting the number of test cases to follow. The description of each test case contains 2 lines. The first of these lines contains a single integer N, the number of cars. The second line contains N space-separated integers, denoting the maximum speed of the cars in the order they entered the long straight segment.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ speed ≤ 10For each test case, output a single line containing the number of cars that were moving at their maximum speed on the segment.Sample Input:
3
1
10
3
8 3 6
5
4 5 1 2 3
Sample Output:
1
2
2, I have written this Solution Code: #include <iostream>
using namespace std;
#define int long long
signed main() {
// your code goes here
int t;
cin>>t;
while(t--){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int x;
cin>>x;
int arr[x];
for(int i=0;i<x;i++){
cin>>arr[i];
}
int count=1;
for(int i=1;i<x;i++){
if(arr[i]<=arr[i-1]){
count++;
}
else{
arr[i]=arr[i-1];
}
}
cout<<count<<endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arun is known for his love of racing sports. He loves Formula 1 with a passion. He traveled to New Delhi to watch the Indian Grand Prix of this year. He noticed that one segment of the circuit was a long straight road. It was impossible for a car to overtake other cars on this segment. Therefore, a car had to lower down its speed if there was a slower car in front of it. While watching the race, Arun started to wonder how many cars were moving at their maximum speed.
Formally, you are provided the max speed of N cars in the sequence in which they entered the circuit's lengthy straightaway. Each car will prefer traveling at its max speed. If the front car's slowness prevents that from happening, it could have to slow down. Besides that, it continues to travel at its max speed while avoiding any crashes. You can assume that the straight section is infinitely long for the purposes of this problem. Count the number of vehicles that were traveling down the straight segment at their max speed.The first line of the input contains a single integer T denoting the number of test cases to follow. The description of each test case contains 2 lines. The first of these lines contains a single integer N, the number of cars. The second line contains N space-separated integers, denoting the maximum speed of the cars in the order they entered the long straight segment.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ speed ≤ 10For each test case, output a single line containing the number of cars that were moving at their maximum speed on the segment.Sample Input:
3
1
10
3
8 3 6
5
4 5 1 2 3
Sample Output:
1
2
2, I have written this Solution Code: import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
class Main {
public static void main(String args[] ) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
int cars=0;
int max=0;
for(int i=0;i<t;i++){
int n=Integer.parseInt (br.readLine());
String[] speeds=br.readLine().split(" ");
// if no of cars ==1
if (n==1){
cars++;
}
else {
max = Integer.parseInt(speeds[0]);
cars++;
for (int j = 0; j < n - 1; j++) {
if (Integer.parseInt(speeds[j + 1]) <= max) {
cars++;
max = Integer.parseInt(speeds[j + 1]);
}
}
}
System.out.println(cars);
cars=0;
}
}
}, In this Programming Language: Java, 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 messages. Here, we will start with the famous <b>"Hello World"</b> message. Now, here you are given a function to complete. <i>Don't worry about the ins and outs of functions, <b>just add the printing command to print "Hello World", </b></i>.your task is to just print "Hello World", without the quotes.Hello WorldHello World must be printed., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
System.out.print("Hello World");
}
}, In this Programming Language: Java, 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 messages. Here, we will start with the famous <b>"Hello World"</b> message. Now, here you are given a function to complete. <i>Don't worry about the ins and outs of functions, <b>just add the printing command to print "Hello World", </b></i>.your task is to just print "Hello World", without the quotes.Hello WorldHello World must be printed., I have written this Solution Code: def print_fun():
print ("Hello World")
def main():
print_fun()
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: def compound_interest(principle, rate, time):
Amount = principle * (pow((1 + rate / 100), time))
CI = Amount - principle
print( '%.2f'%CI)
principle,rate,time=map(int, input().split())
compound_interest(principle,rate,time), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: function calculateCI(P, R, T)
{
let interest = P * (Math.pow(1.0 + R/100.0, T) - 1);
return interest.toFixed(2);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int p,r,t;
cin>>p>>r>>t;
double rate= (float)r/100;
double amt = (float)p*(pow(1+rate,t));
cout << fixed << setprecision(2) << (amt - p);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s= br.readLine().split(" ");
double[] darr = new double[s.length];
for(int i=0;i<s.length;i++){
darr[i] = Double.parseDouble(s[i]);
}
double ans = darr[0]*Math.pow(1+darr[1]/100,darr[2])-darr[0];
System.out.printf("%.2f",ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Being told that an unsorted array contains (n - 1) of n consecutive numbers (where the bounds are defined), find the missing number in O(n) time.
The bounds tell that all the numbers between the lower bound and the upper bound are present in the array except one number which is missing. You have to find that missing number.The function takes three arguments , the first argument is the array of integers, the second argument is the upper bound and the third argument is the lower bound.
All the numbers between the lower and the upper bounds are present in the array (inclusive of both upper and lower bound) except one number which is missing.
Input is provided in the form of an array which would have 3 elements. The first element is the array of integers, the second element is the upper bound and the third element is the lower bound. All three elements are used internally to call the function.
Example: [[1, 4, 3] ,4, 1]
Here, [1,4,3] is the array of integers
4 is the upper bound
1 is the lower boundThe function should print the missing number in the console.const input = [[1, 4, 3] ,4, 1];
const arr = input[0];
const upper_bound = input[1];
const lower_bound = input[2];
findMissingNumber(arr, upper_bound, lower_bound); //prints 2
// Explanation: From numbers 1 to 4, only 2 is missing from the array, I have written this Solution Code:
function findMissingNumber(arrayOfIntegers, upperBound, lowerBound) {
// Iterate through array to find the sum of the numbers
let sumOfIntegers = 0;
for (let i = 0; i < arrayOfIntegers.length; i++) {
sumOfIntegers += arrayOfIntegers[i];
}
// Find theoretical sum of the consecutive numbers using a variation of Gauss Sum.
// Formula: [(N * (N + 1)) / 2] - [(M * (M - 1)) / 2];
// N is the upper bound and M is the lower bound
upperLimitSum = (upperBound * (upperBound + 1)) / 2;
lowerLimitSum = (lowerBound * (lowerBound - 1)) / 2;
theoreticalSum = upperLimitSum - lowerLimitSum;
console.log(theoreticalSum - sumOfIntegers);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<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 <= 100
<b>Note:</b>
<i>But there is a catch here, given user function has already code in it which may or may not be correct, now you need to figure out these and correct them if it is required</i>Print all the even numbers from 1 to n. (print all the numbers in the same line, space-separated)Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: def For_Loop(n):
string = ""
for i in range(1, n+1):
if i % 2 == 0:
string += "%s " % i
return string
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<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 <= 100
<b>Note:</b>
<i>But there is a catch here, given user function has already code in it which may or may not be correct, now you need to figure out these and correct them if it is required</i>Print all the even numbers from 1 to n. (print all the numbers in the same line, space-separated)Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: public static void For_Loop(int n){
for(int i=2;i<=n;i+=2){
System.out.print(i+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
int a[]=new int[str.length];
int sum=0;
for(int i=0;i<str.length;i++)
{
a[i]=Integer.parseInt(str[i]);
sum=sum+a[i];
}
System.out.println(sum/4);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: a,b,c,d = map(int,input().split())
print((a+b+c+d)*100//400), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a list Arr[] of integers representing a list compressed with run-length encoding.
Consider each adjacent pair of elements [freq, val] = [Arr[2*i], Arr[2*i+1]] (with i >= 0). For each such pair, there are freq elements with value val concatenated in a subarray. Concatenate all the subarray from left to right to generate the decompressed array.
Print the decompressed array.The input line contains T, denoting the number of testcases. Each testcase contains 2 lines. The first line contains size of array. Second line contains elements of array separated by space.
Note: size of input array is even
Constraints:
1 <= T <= 100
2 <= N <= 100
0 <= Arr[i] <= 100For each testcase you need to print the decompressed array in a new line.Sample Input:
2
4
1 2 3 4
4
1 1 2 3
Sample Output:
2 4 4 4
1 3 3
Explanation:
Testcase 1: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
At the end the concatenation [2] + [4,4,4] is [2,4,4,4]., I have written this Solution Code: n = int(input())
for _ in range(n):
nt = int(input())
l = [int(x) for x in input().split()];
for i in range(0,nt,2):
print((str(l[i+1]) + " ") * l[i],end="")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a list Arr[] of integers representing a list compressed with run-length encoding.
Consider each adjacent pair of elements [freq, val] = [Arr[2*i], Arr[2*i+1]] (with i >= 0). For each such pair, there are freq elements with value val concatenated in a subarray. Concatenate all the subarray from left to right to generate the decompressed array.
Print the decompressed array.The input line contains T, denoting the number of testcases. Each testcase contains 2 lines. The first line contains size of array. Second line contains elements of array separated by space.
Note: size of input array is even
Constraints:
1 <= T <= 100
2 <= N <= 100
0 <= Arr[i] <= 100For each testcase you need to print the decompressed array in a new line.Sample Input:
2
4
1 2 3 4
4
1 1 2 3
Sample Output:
2 4 4 4
1 3 3
Explanation:
Testcase 1: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
At the end the concatenation [2] + [4,4,4] is [2,4,4,4]., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main(String args[])throws IOException
{
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-- > 0)
{
int N = Integer.parseInt(read.readLine());
String str[] = read.readLine().trim().split(" ");
int arr[] = new int[N];
for(int i = 0; i < N; i++)
arr[i] = Integer.parseInt(str[i]);
int res[] = new int[decompress_EncodedArray(arr, N).length];
res = decompress_EncodedArray(arr, N);
//Collections.addAll(list, );
print(res);
System.out.println();
}
}
static void print(int list[])
{
for(int i = 0; i < list.length; i++)
System.out.print(list[i] + " ");
}
public static int [] decompress_EncodedArray(int[] nums, int n) {
int[] intArr=null;
int opArrSize=0;
int opArrIndex=0;
for(int i=0; i<nums.length; i += 2)
{
opArrSize += nums[i];
}
intArr= new int[opArrSize];
for(int j=0; j<nums.length; j += 2)
{
int freq= nums[j];
int val = nums[j+1];
while(freq>0)
{
intArr[opArrIndex] = val;
freq--;
opArrIndex++;
}
}
return intArr;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a list Arr[] of integers representing a list compressed with run-length encoding.
Consider each adjacent pair of elements [freq, val] = [Arr[2*i], Arr[2*i+1]] (with i >= 0). For each such pair, there are freq elements with value val concatenated in a subarray. Concatenate all the subarray from left to right to generate the decompressed array.
Print the decompressed array.The input line contains T, denoting the number of testcases. Each testcase contains 2 lines. The first line contains size of array. Second line contains elements of array separated by space.
Note: size of input array is even
Constraints:
1 <= T <= 100
2 <= N <= 100
0 <= Arr[i] <= 100For each testcase you need to print the decompressed array in a new line.Sample Input:
2
4
1 2 3 4
4
1 1 2 3
Sample Output:
2 4 4 4
1 3 3
Explanation:
Testcase 1: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
At the end the concatenation [2] + [4,4,4] is [2,4,4,4]., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i+=2){
while(a[i]>0){
cout<<a[i+1]<<" ";
a[i]--;
}
}
cout<<endl;}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of n pairs of integers. Your task is to sort the array on the basis of the first element of pairs in descending order. If the first element is equal in two or more pairs then give preference to the pair that has a greater second element value.<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>SortPair()</b> that takes the array of pairs and the integer N(size of the array) as a parameter.
<b>Custom Input <b/>
The first line of input will be a single integer N. The next line of input contains 2*N space-separated integers where unique adjacent elements are pairs. Custom input for 1st sample:-
4
1 2 3 4 5 6 7 8
<b>Constraints:-</b>
1<=N<=10<sup>3</sup>
1<=value<=10<sup>5</sup>Return the sorted array of pairs.Sample Input 1:
4
(1, 2), (3, 4), (5, 6), (7, 8)
Sample Output 1:
(7, 8), (5, 6), (3, 4), (1, 2)
Sample Input 2:
3
(1, 1), (2, 2), (3, 3)
Sample Output 2:
(3, 3), (2, 2), (1, 1)
Sample Input 3:
3
(1, 1), (1, 2), (3, 3)
Sample Output 3:
(3, 3), (1, 2), (1, 1)
<b>Explanation :</b>
(1,2) and (1,1) have the same first element. But (1,2) has a greater second element so (1,2) comes before (1,1) in a sorted array.
, I have written this Solution Code: def SortPair(items,n):
items.sort(reverse = True)
return items, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of n pairs of integers. Your task is to sort the array on the basis of the first element of pairs in descending order. If the first element is equal in two or more pairs then give preference to the pair that has a greater second element value.<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>SortPair()</b> that takes the array of pairs and the integer N(size of the array) as a parameter.
<b>Custom Input <b/>
The first line of input will be a single integer N. The next line of input contains 2*N space-separated integers where unique adjacent elements are pairs. Custom input for 1st sample:-
4
1 2 3 4 5 6 7 8
<b>Constraints:-</b>
1<=N<=10<sup>3</sup>
1<=value<=10<sup>5</sup>Return the sorted array of pairs.Sample Input 1:
4
(1, 2), (3, 4), (5, 6), (7, 8)
Sample Output 1:
(7, 8), (5, 6), (3, 4), (1, 2)
Sample Input 2:
3
(1, 1), (2, 2), (3, 3)
Sample Output 2:
(3, 3), (2, 2), (1, 1)
Sample Input 3:
3
(1, 1), (1, 2), (3, 3)
Sample Output 3:
(3, 3), (1, 2), (1, 1)
<b>Explanation :</b>
(1,2) and (1,1) have the same first element. But (1,2) has a greater second element so (1,2) comes before (1,1) in a sorted array.
, I have written this Solution Code:
static Pair[] SortPair(Pair arr[], int n)
{
// Comparator to sort the pair according to second element
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
if(p1.x==p2.x){
return p2.y-p1.y;
}
return p2.x-p1.x;
}
});
return arr;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to calculate values for each of the following operations:-
a + b
a - b
a * b
a/bSince this will be a functional problem, you don't have to take input. You have to complete the function
<b>operations()</b> that takes the integer a and b as parameters.
<b>Constraints:</b>
1 ≤ b ≤ a ≤1000
<b> It is guaranteed that a will be divisible by b.</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5, I have written this Solution Code: def operations(x, y):
print(x+y)
print(x-y)
print(x*y)
print(x//y), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
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));
String str1 = br.readLine();
String str2[] = str1.split(" ");
int n = Integer.parseInt(str2[0]);
int k = Integer.parseInt(str2[1]);
String str3 = br.readLine();
String str4[] = str3.split(" ");
long[] arr = new long[n];
for(int i = 0; i < n; ++i) {
arr[i] = Long.parseLong(str4[i]);
}
Arrays.sort(arr);
int i=0,j=2;
long ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((arr[j]-arr[i])>k){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: n, p = input().split(' ')
n = int(n)
p = int(p)
arr = input().split(' ')[:n]
for i in range(n):
arr[i] = int(arr[i])
arr.sort()
i = 0
k = 2
count = 0
while k!=n:
if i == k-1:
k = k + 1
continue
if (arr[k] - arr[i]) <= p:
count = count + (int)(((k-i)*(k-i-1))/2)
k = k + 1
else:
i = i + 1
print(count)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n,p;
cin>>n>>p;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
int i=0,j=2;
int ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((a[j]-a[i])>p){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was playing with her array A consisting of N integers, her small sister Tono asked her a problem.
Given the array A, you can can perform the following operation at max K times:
<li> Choose two alternate indices {i, i+1} (1 <= i <= N-1), make A[i] = A[i] + A[i+1] <b>OR</b> A[i+1] = A[i] + A[i+1].
Let's assume the minimum value in the array is M after K operations. Find the maximum possible value of M.
As Solo is small, please solve the problem for her.The first line of the input contains two integers N and K.
The second line of the input contains N space separated integers, the elements of array A.
Constraints
2 <= N <= 10000
1 <= K <= 1000000
1 <= A[i] <= 100000000Output a single integer, the answer to the problem modulo 1000000007.Sample Input
5 5
8 1 2 3 4
Sample Output
9
Explanation: We perform the following operations:
- Make A[1] = A[1] + A[2]
- Make A[2] = A[1] + A[2]
- Make A[3] = A[2] + A[3]
- Make A[4] = A[3] + A[4]
- Make A[5] = A[4] + A[5]
Final array = [9, 10, 12, 15, 19]. Minimum value = 9., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int t[10005];
signed main() {
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n, k; cin>>n>>k;
for (int i = 0; i < n; ++i) {
cin >> t[i];
}
if (k < n) {
nth_element(t, t + k, t + n);
cout << t[k] << endl;
return 0;
}
k -= n - 1;
int a = 1;
int b = 0;
for (int i = 0; i < k; ++i) {
a += b;
b = a - b;
if (a > (int) 1e9) {
break;
}
}
pii best(0LL, 0LL);
for (int i = 0; i < n - 1; ++i) {
int x = max(t[i], t[i + 1]);
int y = min(t[i], t[i + 1]);
if (a * x + b * y > a * best.F + b * best.S) {
best = {x, y};
}
}
int x = best.F;
int y = best.S;
for (int i = 0; i < k; ++i) {
x += y;
y = x - y;
x %= MOD;
}
cout << x << endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon or not.<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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code:
int DragonSlayer(int A, int B, int C,int D){
int x = C/B;
if(C%B!=0){x++;}
int y = A/D;
if(A%D!=0){y++;}
if(x<y){return 0;}
return 1;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon or not.<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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code: static int DragonSlayer(int A, int B, int C,int D){
int x = C/B;
if(C%B!=0){x++;}
int y = A/D;
if(A%D!=0){y++;}
if(x<y){return 0;}
return 1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |