Codeforces Round #319 (Div. 2) 577B Modulo Sum(dp)

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.

给你n个数字和一个m，问你n个数字中任意个数之和是否能整除m。

首先判断是否n大于等于m，如果是的话就一定可以整除m。如果不是就转化为01背包问题(暴搜一定TLE)，读入数据的时候直接取模，dp[i][j]的含义

是“加上a[i]对m去模为j”，状态转移方程：dp[i][(j + a[i]) % m] = 1

AC代码：

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int MAXN = 2020;int a[MAXN], dp[MAXN][MAXN], n, m;bool flag;int main(int argc, char const *argv[]){scanf("%d%d", &n, &m);if(n >= m) {for(int i = 0; i < n; ++i)scanf("%d", &m);printf("YES\n");return 0;}for(int i = 0; i < n; ++i)scanf("%d", &a[i]);for(int i = 0; i < n; ++i) {if(!i) dp[0][(a[i] % m)] = 1;else {dp[i][(a[i] % m)] = 1;for(int j = 0; j < m; ++j) if(dp[i - 1][j]) {dp[i][(j + a[i]) % m] = 1;dp[i][j] = dp[i - 1][j];}}}for(int i = 0; i < n; ++i)if(dp[i][0]) {flag = true;break;}if(flag) printf("YES\n");else printf("NO\n");return 0;}