CF577B Modulo Sum 好题

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 by6, 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：

给出一个数列，和一个数m，问能不能从这个数列中选出若干个数，使得这些数的和可以整除m

整除m，也就是%m==0

其实是个很水的01背包，每个数取和不取

dp[i][j]表示选择到第i个数，和模m==j的情况有没有(有1，没有0)

但是我们会发现n很大，m很小

根据抽屉原理，当n>=m时，一定能

当n<m时，此时的数据大小<=1000,这个时候的dp，复杂度为n*m<m*m，可以过了

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxm=1e3+10;
const int maxn=1e6+10;

int dp[maxm][maxm];
int a[maxn];

int main()
{
    int n,m;

    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
        a[i]%=m;
    }
    if(n>=m){
        printf("YES
");
        return 0;
    }

    memset(dp,0,sizeof dp);

    for(int i=1;i<=n;i++){
        dp[i][a[i]]=1;
        for(int j=0;j<m;j++){
            if(dp[i-1][j]){
                dp[i][j]=true;
                dp[i][(j+a[i])%m]=true;
            }
        }
    }

    if(dp[n][0]>0)
        printf("YES
");
    else
        printf("NO
");

    return 0;
}