BestCoder Round #85 hdoj5776 sum

最新推荐文章于 2018-04-01 14:04:18 发布

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本文介绍了一种算法来判断一个整数序列中是否存在连续子序列，其和能被特定整数m整除。通过预处理前缀和并运用抽屉原理，文章提供了一个高效的解决方案。

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Problem Description

Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO

Input

The first line of the input has an integer T ( $\leq T \leq 10$ ), which represents the number of test cases. For each test case, there are two lines: 1.The first line contains two positive integers n, m ( $\leq n \leq 100000$ , $\leq m \leq 5000$ ). 2.The second line contains n positive integers x ( $\leq x \leq 100$ ) according to the sequence.

Output

Output T lines, each line print a YES or NO.

Sample Input

Sample Output

YES
NO

思路：先预处理前缀和，就是把前1项的和，前两项的和.....存到一个数组中，然后让这个数组中的每一个元素对m取余，如果这个取余的结果出现了两次，那么就输出YES.利用抽屉原理，如果n大于m，则输出YES

代码如下：

#include<cstdio>
#include<cstring>
int sum[100010];
int a[100010];
int book[5050];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m,i,flag=1;
		memset(book,0,sizeof(book));
		book[0]=1;
		scanf("%d%d",&n,&m);
	
		scanf("%d",&a[0]);
		sum[0]=a[0];
        for(i=1;i<n;i++)
		{
		  scanf("%d",&a[i]);
		  sum[i]=sum[i-1]+a[i];
		}
		
		//printf("%d\n",a[0]);
		for(i=0;i<n;i++)
		{
			int l=sum[i]%m;
            if(book[l]==1)
			{
				flag=0;
				break;
			 } 
			 book[l]=1;  
		}
		if(flag)
		printf("NO\n");
		else
		printf("YES\n");
		
	}
}