Codeforces 577B Modulo Sum 动态规划基础

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, ..., a_n, and a numberm.

Check if it is possible to choose a non-empty subsequence a_ij such that the sum of numbers in this subsequence is divisible bym.

Input

The first line contains two numbers, n andm (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, ..., a_n (0 ≤ a_i ≤ 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 and3, the sum of which is divisible by 5.

In the second sample test the single non-empty subsequence of numbers is a single number5. 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：给你一组数，从中找出一些数，求和是m的倍数。

题解：

Let's consider two cases: n > m andn ≤ m.

If n > m, let's look at prefix sums. By pigeonhole principle, there are two equals sums modulom. Assume S_lmodm = S_rmodm. Then the sum on segment[l + 1, r] equals zero modulo m, that means the answer is definitely "YES".

If n ≤ m, we'll solve this task using dynamic programming inO(m²) time. Assumecan[i][r] means if we can achieve the sum equal tor modulo m using only firsti - 1 items. The updates in this dynamic programming are obvious: we either take numbera_i and go to the statecan[i + 1][(r + a_i)mod m] or not, then we'll get to the state can[i + 1][r].

The complexity is O(m²).

分析：n>m,鸽巢原理，一定可以

n<=m,从一组数中选择选择一些数，求和看看是否满足某个条件，变形的0/1背包问题，可以用动态规划解决

#include 
using namespace std;
int dp[1010][1010], a[100010], n, m;
int main()
{
    scanf("%d%d",&n,&m);
    for (int i = 1; i <= n; i ++) {
        scanf("%d",&a[i]);
        a[i] %= m;
    }
    if (n > m) {
        printf("YES\n");
        return 0;
    }
    for (int i = 0; i <= n; i ++)
        dp[i][0] = true;
    for (int i = 1; i <= n; i ++)
        for (int j = 0; j <= m; j ++)
            dp[i][j] = dp[i-1][j] || dp[i-1][(j + a[i]) % m];
    printf("%s\n", dp[n][m] ? "YES" : "NO");
   return 0;
}