cf 577B

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.

Sample test(s)

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

Note

In the first sample test you can choose numbers 2 and 3,
the sum of which is divisible by 5.

In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is
not divisible by 6, that is, the sought subsequence doesn't exist.

In the third sample test you need to choose two numbers 3 on the ends.

In the fourth sample test you can take the whole subsequence.

题意：从n个数里找任意个的和使其能被m(<=1000)整除。

思路：当n>m时，前m+1个前缀和肯定会出现%m余数相同的情况，那么这两个前缀和的差便能被m整除。

当n<=m便直接用dp求得前i个数所能得出的余数便可。时间复杂度（m*m）。

#pragma comment(linker, "/STACK:102400000000,102400000000")
#include<iostream>
#include<stdio.h>
#include<math.h>
#include <string>
#include<string.h>
#include<map>
#include<queue>
#include<set>
#include<utility>
#include<vector>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define eps 1e-8
#define pii pair<int,int>
#define inf 0x3f3f3f3f
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define ll long long int
#define mod 1000000007
#define maxn 1005
#define maxm 1000005
int f[maxn][2],a[maxm];
int main()
{
    int n,m;
    rd2(n,m);
    for(int i=1;i<=n;i++) rd(a[i]);
    memset(f,0,sizeof(f));
    if(n>m) printf("YES\n");
    else{
        int pr=0;
        int nt=1;
        for(int i=1;i<=n;i++){
                nt^=1;pr^=1;
            for(int j=0;j<m;j++){
                    f[a[i]%m][nt]=1;
                    if(f[j][pr]) f[(j+a[i])%m][nt]=f[j][nt]=1;
            }
        }
        if(f[0][nt]) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}