Hdu5776 sum 抽屉原理+同余定理

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
  

Source

BestCoder Round #85

Recommend

wange2014

用前缀和去处理就好了，用 a[ i ] 去记录前 i 项之和对 m 的模。那么只要有两项是相同的，就存在所要求的连续串。

将 a[ 0 ]设为0可避免连续串从第一项开始的情况。

此外，由抽屉原理可得，n >= m 时一定存在。

#include <iostream>
#include <cstdio>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;

int main()
{
    int t;
    int n,m;
    int a[100005];
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        int t;
        a[0]=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&t);
            a[i]=(a[i-1]+t)%m;
        }
        if(n>=m){
            printf("YES\n");
        }else{
            sort(a+1,a+n+1);
            int flag=1;
            for(int i=1;i<=n;i++){
                if(a[i]==a[i-1]){
                    flag=0;
                    break;
                }
            }
            if(!flag){
                printf("YES\n");
            }else{
                printf("NO\n");
            }
        }
    }
    return 0;
}