hdu5776 sum

sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 588    Accepted Submission(s): 286

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 (1≤T≤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 (1≤n≤100000, 1≤m≤5000).
2.The second line contains n positive integers x (1≤x≤100) according to the sequence.

 

Output

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

 

Sample Input

2
3 3
1 2 3
5 7
6 6 6 6 6

 

Sample Output

YES
NO

 

若 x % m = b 且 y % m = b，那么x可以写成x = a1 * m + b，y可以写成y = a2 * m + b，(y - x) % m = ((a2 - a1) * m) % m = 0

可以用一个前缀和记录x和y，若出现x % m == y % m，那么x到y这一段的子列和就是m的倍数，由于x % m 在[0, m - 1]之间，

再根据鸽笼原理可知，出现模m后相等的情况不会超过m次求和判断，可求

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <map>

using namespace std;

int T, n, m, cur, sum;
map<int, int> mp;

int main()
{
	cin >> T;
	while (T--) {
		mp.clear();
		scanf("%d%d", &n, &m);
		sum = 0;
		bool flag = false;
		for (int i = 0; i < n; i++) {
			scanf("%d", &cur);
			sum = (sum + cur) % m;
			if (sum == 0) {
				flag = true;
			}
			if (mp[sum] == 0) {
				mp[sum] = 1;
			}
			else {
				flag = true;
			}
		}
		puts(flag ? "YES" : "NO");
	}
	return 0;
}