TOJ 1721 Partial Sums

Description

Given a series of n numbers a1, a2, …, an, the partial sum of the numbers is defined as the sum of ai, ai+1, …, aj.

You are supposed to calculate how many partial sums of a given series of numbers could be divided evenly by a given number m.

Input

There are multiple test, each contains 2 lines.

The first line is 2 positive integers n (n <= 10000 ) and m (m <= 5000).

The second line contains n non-negative integers a1, a2, …, an. Numbers are separated by one or several spaces.

The input is ended by EOF.

Output

One test each line – the number of partial sums which could be divided by m.

Sample Input

5 4
1 2 3 4 5
6 7
9 8 7 6 5 4

Sample Output

2
3

Source

ZOJ Monthly March 2003

 #include <stdio.h>
 int main(int argc, char *argv[])
 {
     int n,m;
     int a[];
     int sum[];
     while( scanf("%d %d" ,&n ,&m)!=EOF ){
         int ans=;
         int ca[]={};
         for(int i=; i<n; i++){
             scanf("%d",&a[i]);
             if(i==){
                 sum[i]=a[i]%m;
             }else{
                 sum[i]=(sum[i-]+a[i])%m;
             }
             if(sum[i]%m==)
                 ans++;
             ca[sum[i]]++;
         }
         for(int i=; i<m; i++)
             ans+=ca[i]*(ca[i]-)/;
         printf("%d\n",ans);
     }
     return ;
 }