Codeforces - 577B 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 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 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：给你一个序列，询问是否存在一个子序列，其中子序列中各个元素能被m整除；

题解：当n > m ，每个元素对m取余，即有n个0~m-1的数，因为n>m，定能找到一个子序列满足条件；‘

若n <= m，即maxn <= 10^3 ；

设dp[i][j] 前i-1个数组成子序列的和被m取余的值为j，存不存在；

代码：

#include <stdio.h>
#define ll __int64
const int maxn=1000010;
const int maxm=1005;
int n, m, i, j, a[maxn], dp[maxm][maxm];
int main()
{
	
	scanf("%d%d", &n, &m);
	if(n > m)
	{
		printf("YES\n");
		return 0;	
	}
	for(i = 1;i <= n;i++)
	{
		scanf("%d", &a[i]);
	}
	for(i = 1;i <= n;i++)
	{
		dp[i+1][a[i]%m] = 1;
		for(j = 0;j < m;j++)
		{
			if(dp[i][j])
			{
				dp[i+1][j] = 1;
				dp[i+1][(j+a[i])%m] = 1;
			}
		}
		
	}
	if(dp[n+1][0]) printf("YES\n");
	else printf("NO\n");
	
}