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[10001];
    int sum[10001];
    while( scanf("%d %d" ,&n ,&m)!=EOF ){
        int ans=0;
        int ca[5001]={0};
        for(int i=0; i<n; i++){
            scanf("%d",&a[i]);
            if(i==0){
                sum[i]=a[i]%m;
            }else{
                sum[i]=(sum[i-1]+a[i])%m;
            }
            if(sum[i]%m==0)
                ans++;
            ca[sum[i]]++;
        }
        for(int i=0; i<m; i++)
            ans+=ca[i]*(ca[i]-1)/2;
        printf("%d
",ans);
    }
    return 0;
}