HDU 5776 sum（抽屉原理）

题目传送：http://acm.hdu.edu.cn/showproblem.php?pid=5776

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

 

直接拿这道代码改了一下，两道题一个意思http://www.cnblogs.com/Annetree/p/7095096.html

 

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<string.h>
 #include<cmath>
 using namespace std;
 #define N 100005

 struct node
 {
     int num,r;
 }k[N];

 bool cmp(node a,node b)
 {
     if(a.r==b.r)
         return a.num<b.num;
     return a.r<b.r;
 }

 int main()
 {
     long long sum,a;
     int n,m;
     bool flag;
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         flag=false;
         if(n>m)
             flag=true;
         sum=;
         k[].num=;
         for(int i=;i<=n;i++)
         {
             scanf("%lld",&a);
             sum+=a;
             k[i].r=sum%m;
             k[i].num=i;
             if(k[i].r==&&flag==false)
             {
                 flag=true;
             }
         }
         if(!flag)
         {
             sort(k+,k++n,cmp);
             for(int i=;i<=n;i++)
             {
                 if(k[i].r==k[i-].r)
                 {
                     flag=true;
                     break;
                 }
             }
         }
         if(!flag)
             printf("NO\n");
         else
             printf("YES\n");
     }
     return ;
 }