Problem:
Score: 300 points

Problem Statement

AtCoder Mart sells 1000000 of each of the six items below:

Riceballs, priced at 100 yen (the currency of Japan) each

Sandwiches, priced at 101 yen each

Cookies, priced at 102 yen each

Cakes, priced at 103 yen each

Candies, priced at 104 yen each

Computers, priced at 105 yen each

Takahashi wants to buy some of them that cost exactly X yen in total.
Determine whether this is possible.

(Ignore consumption tax.)

Constraints

1 \leq X \leq 100000

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

If it is possible to buy some set of items that cost exactly X yen in total, print 1; otherwise, print 0.

Sample Input 1

615

Sample Output 1

1

For example, we can buy one of each kind of item, which will cost 100+101+102+103+104+105=615 yen in total.

Sample Input 2

217

Sample Output 2

0

No set of items costs 217 yen in total.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define REP(i,n) for(long long i=0;i<n;i++)
#define REP1(i,n) for(long long i=1;i<=n;i++)
#define REP2D(i,j,h,w) for(long long i=0;i<h;i++) for(long long j=0;j<w;j++)
#define REP2D1(i,j,h,w) for(long long i=1;i<=h;i++) for(long long j=1;j<=w;j++)
#define PER(i,n) for(long long i=((n)-1);i>=0;i--)
#define PER1(i,n) for(long long i=(n);i>0;i--)
#define FOR(i,a,b) for(long long i=(a);i<(b);i++)
#define FORE(i,a,b) for(long long i=(a);i<=(b);i++)
#define ITE(arr) for(auto ite=(arr).begin();ite!=(arr).end();++ite)
#define ALL(a) (a.begin()),(a.end())
#define RANGE(a) (a),(a+sizeof(a))
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define YNPRT(b) cout<<((b)?"Yes":"No")<<endl
#define ENTER printf("\n")
#define REV(arr) reverse(ALL(arr))
#define PRT(a) cout<<a<<endl
#ifdef DEBUG
#define DBPRT(a) cout << "[Debug] - " << #a << " : " << a << endl
#define DBSTART if(1){
#define DBEND }
#else
#define DBPRT(a) do{}while(0)
#define DBSTART if(0){
#define DBEND }
#endif
#define PRTLST(arr,num) REP(_i,num) cout<<_i<<" - "<<arr[_i]<<endl;
#define PRTLST2(arr2,d1,d2) REP(_i,d1) REP(_j,d2) cout<<_i<<","<<_j<<" : "<<arr2[_i][_j]<<endl;
#define PRTLST2D(arr2,d1,d2) do{cout<<"L\t";REP(_i,d2) cout<<_i<<"\t"; cout<<endl; REP(_i,d1){cout<<_i<<"\t";REP(_j,d2){cout<<arr2[_i][_j]<<"\t";}cout<<endl;}}while(0);

#define TOSUM(arr,sum,n) {sum[0]=arr[0];REP1(i,n-1) sum[i]=sum[i-1]+arr[i];}

#define MIN(target,v1) (target)=min(target,v1)
#define MAX(target,v1) (target)=max(target,v1) 
#define P1 first
#define P2 second
#define PB push_back
#define UB upper_bound
#define LB lower_bound
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const int INF_INT = 2147483647;
const ll INF_LL = 9223372036854775807LL;
const ull INF_ULL = 18446744073709551615Ull;
const ll P = 92540646808111039LL;
const int Move[4][2] = {-1,0,0,1,1,0,0,-1};
const int Move_[8][2] = {-1,0,-1,-1,0,1,1,1,1,0,1,-1,0,-1,-1,-1};
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const pair<T1,T2>& p) {os<<"( "<<p.P1<<" , "<<p.P2<<" )";return os;}
template<typename T>
ostream& operator<<(ostream& os,const vector<T>& v) {ITE(v) os << (ite-v.begin()) << " : " << *ite <<endl;return os;}
template<typename T>
ostream& operator<<(ostream& os,const set<T>& v) {os<<" { ";ITE(v) {os<<*ite;if(ite!=--v.end()){os<<" , ";}} os<<" } ";return os;}
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const map<T1,T2>& m) {ITE(m) {os<<ite->P1<<"\t\t|->\t\t"<<ite->P2<<endl;} return os;}

//---------------------
#define MAXN 100005
//---------------------

ll x ;
ll mn[6];
bool dp[6][MAXN];
int main(){
	cin >> x;
	REP(i,6) mn[i] = 100 + i;
	ZERO(dp);
	REP(i,x+1) dp[0][i] = (i%mn[0] == 0);
	REP(i,6) dp[i][0] = true;
	REP1(i,5) REP1(j,x) {
		if(j-mn[i]>=0) dp[i][j] = dp[i-1][j] || dp[i][j-mn[i]];
		else dp[i][j] = dp[i-1][j];
	}
	if(dp[5][x]) PRT(1);
	else PRT(0);
	return 0;
}