CF #319 B Modulo Sum

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
#include<set>
#include<bitset>
//#define ONLINE_JUDGE
#define eps 1e-5
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
#define ll __int64
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
//#pragma comment(linker,"/STACK:1024000000,1024000000")
int n,m;
int p;
#define M 1110
#define N 1000010
#define Mod 1000000000
#define p(x,y) make_pair(x,y)
const int MAX_len=550;
int isp[M];
vector<int> prime;
int dp[1010][2];
int a
;
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
#endif
while(sfd(n,m)!=EOF){
int flag=0;
for(int i=0;i<n;i++){
sf(a[i]);
a[i] %= m;
if(a[i] == 0)
flag=1;
}
if(flag){
printf("YES\n");
continue;
}
int p=1;
memset(dp,0,sizeof dp);
for(int i=0;i<n;i++){
dp[a[i]][p] = 1;        //能够到达a[i]
for(int j=0;j<m;j++){   //循环所有可能的对m取余的结果
if(dp[j][p^1]){     //如果j这个结果能够在前面取到
dp[(a[i]+j)%m][p] = 1;  //那么到了a[i],我们可以取a[i]，余数就变成了(a[i]+j)%m
dp[j][p] = 1;           //也可以不取a[i],余数就还是 保持j不变
}
}
if(dp[0][p]){   //前p个数能够取到对m取余为0的和结果
flag=1;
break;
}
p ^= 1;
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}