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≤101≤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≤1000001≤n≤100000, 1≤m≤50001≤m≤5000).
2.The second line contains n positive integers x (1≤x≤1001≤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
现在假设有一个正整数序列a1,a2,a3,a4……an,试证明我们一定能够找到一段连续的序列和,让这个和是n的倍数,该命题的证明就用到了抽屉原理
我们可以先构造一个序列si=a1+a2+…ai
然后分别对于si取模,如果其中有一个sk%n==0,那么a1+a2+…+ak就一定是n的倍数(该种情况得证)
下面是上一种情况的反面,即任何一个sk对于n的余数都不为0
对于这种情况,我们可以如下考虑,因为si%n!=0
那么si%n的范围必然在1——(n-1),所以原序列si就产生了n个范围在1——(n-1)的余数,于是抽屉原理就来了,n个数放进n-1个盒子里面,必然至少有两个余数会重复,那么这两个sk1,sk2之差必然是n的倍数,
而sk1-sk2是一段连续的序列,那么原命题就得到了证明了
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
int n, m, sum, num, flag = 0;
bool vis[5005];
scanf("%d%d", &n, &m);
memset(vis, false, sizeof(vis));
sum = 0;
vis[0] = 1;
for(int i = 1; i <= n; i++)
{
scanf("%d", &num);
sum += num;
if(vis[sum%m]) flag = 1;
vis[sum%m] = true;
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}