Codeforces Round #319 (Div. 2)B. Modulo Sum DP

                                                         B. Modulo Sum

                                                                                                 time limit per test

                                                                                                 2 seconds

                                                                                                 memory limit per test

                                                                                                 256 megabytes

 

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

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：给你n,m，n个数，让你从中找出任意数的和mod M==0

题解：背包dp

//1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define memfy(a) memset(a,-1,sizeof(a))
#define TS printf("111111
");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define maxn 1000005
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//****************************************
int hashs[maxn],a[maxn*3];
bool  dp[3][maxn];
int main()
{

    mem(hashs);
    int flag=0;
    int n=read(),m=read();
    FOR(i,1,n)
    {
        scanf("%d",&a[i]);
       a[i]=a[i]%m;
       if(a[i]==0)a[i]=m;
    }dp[0][0]=true;
    dp[1][0]=true;
    for(int i=1;i<=n;i++)
    {
        for(int j=m;j>=0;j--)
        {
            if(dp[1][j])
            {
                if(j+a[i]==m){
                    puts("YES");
                    return 0;
                }
                dp[0][(j+a[i])%m]=true;
            }
        }
        memcpy(dp[1],dp[0],sizeof(dp[0]));
    }
    cout<<"NO"<<endl;
    return 0;
}