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

借鉴自http://blog.csdn.net/u014492306/article/details/48369857

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 51 2 3

output

YES

input

1 65

output

NO

input

4 63 1 1 3

output

YES

input

6 65 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.

借鉴了题解思路。

当n>=m时，抽屉原理即可知必定存在子列可整除m。

当n<m时，由于m范围较小，可直接计算n个数字所能产生的所有余数，最后检测0是否存在即可。代码中为了防止重复将本身计算，用了一些小技巧，值得学习

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdlib>#include <string>#include <vector>#include <cstdio>#include <stack>#include <cmath>#include <queue>#include <map>#include <set>using namespace std;#define N 1000005#define INF 0x3f3f3f3f;int a[N],f[1005][2];int main() {    int n,m;        while (cin>>n>>m) {        for (int i=0; i<n; i++) {            scanf("%d",&a[i]);        }        memset(f, 0, sizeof(f));        if (n>=m) {            cout<<"YES"<<endl;        }        else {            int pr=0,nt=1;            for (int i=0; i<n; i++) {                nt^=1;                pr^=1;                f[a[i]%m][nt]=1;                for (int j=0; j<m; j++) {                    if (f[j][pr]) {                        f[(j+a[i])%m][nt]=f[j][nt]=1;                    }                }            }        if (f[0][nt]==1) {            cout<<"YES"<<endl;        }        else cout<<"NO"<<endl;        }    }    return 0;}

。