codeforces 577B B. Modulo Sum(水题)

题目链接：

B. Modulo Sum

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequence of numbers a1, a2, ..., an, and a number m.

Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.

Input

The first line contains two numbers, n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.

Examples

input

3 5
1 2 3

output

YES

input

1 6
5

output

NO

input

4 6
3 1 1 3

output

YES

input

6 6
5 5 5 5 5 5

output

YES

题意：

能否找到一个子序列的和能被m整除;

思路：

有一道hdu的水题就是这样的,以前傻傻的居然不会做;就是把余数拿出来搞搞就好了;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
  
using namespace std;
  
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
  
typedef  long long LL;
  
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
  
const int mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e6+20;
const int maxn=1e3+110;
const double eps=1e-12;
 
int n,m,a[N],vis[maxn],temp[maxn];
int main()
{
    read(n);read(m);
    int flag=0;
    For(i,1,n)read(a[i]);
    vis[0]=1;
    For(i,1,n)
    {
        a[i]%=m;
        int t=(m-a[i])%m;
        if(vis[t]){flag=1;break;}
        mst(temp,0);
        for(int j=0;j<m;j++)
        {
            if(vis[j])temp[(j+a[i])%m]=1;
        }
        for(int j=0;j<m;j++)
            if(temp[j])vis[j]=1;
    }
    if(flag)cout<<"YES
";
    else cout<<"NO
";
    return 0;
}