Codeforces #319(Div.2) B. Modulo Sum （动态规划）

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.

题意：两个数n和m，还有a1-an的n个数，判断是否存在该数列的一个子序列，使得子序列元素的和可以被m整除。

看了下官方题解：

分两种情况：n>m 和 n<=m。

如果n>m, 可以判断输出一定为“Yes”。求出前 i 个数的和 S1-Sn，有鸽巢原理，可以知道至少有两个数列和对m取模的结果相等，假设为Sl%m=Sr%m，则可以知道（Sl-Sr）%m==0，于是 [ l+1,r ]就是所求的子序列。

如果n<=m, 用动态规划解决，O(m^2)。dp[i][r]表示到了第i个数，前面子序列的和对m取模是否能够等于r，于是状态转移方程就是，如果前面一个数，有dp[i-1][r]， 则后面一个数可以选择ai使得dp[i][(r+ai)%m]为1，或者不选ai，使dp[i][r]=1。这样做的目的是为了尽最大努力使得后面的数能够得到对m取模为某个数的子序列和。遍历一遍dp[i][0]就得到本题的答案。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1010;int dp[maxn][maxn], a[maxn];int main(){int n, m;while (scanf("%d%d", &n, &m) != EOF){if (n >= m){for (int i = 0; i < n; i++)scanf("%d", &m);printf("YES\n");continue;}for (int i = 0; i < n; i++)scanf("%d", &a[i]);memset(dp, 0, sizeof(dp));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];}}}}int flag = 0;for (int i = 0; i < n; i++){if (dp[i][0])flag = 1;}if (flag)printf("YES\n");elseprintf("NO\n");}}