poj 3844 Divisible Subsequences

Description

Given a sequence of positive integers, count all contiguous subsequences (sometimes called substrings, in contrast to subsequences, which may leave out elements) the sum of which is divisible by a given number. These subsequences may overlap. For example, the sequence (see sample input) 
2, 1, 2, 1, 1, 2, 1, 2
contains six contiguous subsequences the sum of which is divisible by four: the first to eighth number, the second to fourth number, the second to seventh number, the third to fifth number, the fourth to sixth number, and the fifth to seventh number.

Input

The first line of the input consists of an integer c (1 <= c <= 200), the number of test cases. Then follow two lines per test case. 
Each test case starts with a line consisting of two integers d (1 <= d <= 1 000 000) and n (1 <= n <= 50 000), the divisor of the sum of the subsequences and the length of the sequence, respectively. The second line of a test case contains the elements of the sequence, which are integers between 1 and 1 000 000 000, inclusively.

Output

For each test case, print a single line consisting of a single integer, the number of contiguous subsequences the sum of which is divisible by d.

Sample Input

27 31 2 34 82 1 2 1 1 2 1 2

Sample Output

06

Source

2009 ACM North Western European Regional Contest

先求累加和sum[i](a[0]+a[1]+...+a[i])，然后求余，若两个累加和sum[i]%d==sum[j]%d，那么(sum[j]-sum[i])%d==0，即a[i+1], ... , a[j]就是一个满足条件的连续子序列。注意余数为0的情况，可以在原序列最前面加上一个假设的元素0，这就就可以统一处理了。

#include <stdio.h>#include <memory.h>#define MAX_LEN 50010#define MAX_NUMBER 1000010int sum[MAX_LEN];__int64 mod[MAX_NUMBER];int main(){int c;scanf("%d", &c);while(c--){int d, n;scanf("%d%d", &d, &n);int x;memset(mod, 0, sizeof(mod));sum[0] = 0;mod[0] = 1;for(int i = 1; i <= n; i++){scanf("%d", &x);sum[i] = sum[i-1] + x;sum[i] %= d;mod[sum[i]]++;}__int64 result = 0;for(int i = 0; i < d; i++)if(mod[i] > 0)result += mod[i] * (mod[i]-1) / 2;printf("%I64d\n", result);}return 0;}