Light oj 1134 - Be Efficient

题目链接：http://lightoj.com/volume_showproblem.php?problem=1134

1134 - Be Efficient

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Time Limit: 2 second(s)Memory Limit: 32 MB

You are given an array with N integers, and another integer M. You have to find the number of consecutive subsequences which are divisible by M.

For example, let N = 4, the array contains {2, 1, 4, 3} and M = 4.

The consecutive subsequences are {2}, {2 1}, {2 1 4}, {2 1 4 3}, {1}, {1 4}, {1 4 3}, {4}, {4 3} and {3}. Of these 10 'consecutive subsequences', only two of them adds up to a figure that is a multiple of 4 - {1 4 3} and {4}.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case contains two integers N (1 ≤ N ≤ 10⁵) and M (1 ≤ M ≤ 10⁵). The next line contains N space separated integers forming the array. Each of these integers will lie in the range [1, 10⁵].

Output

For each case, print the case number and the total number of consecutive subsequences that are divisible by M.

Sample Input

Output for Sample Input

2

4 4

2 1 4 3

6 3

1 2 3 4 5 6

Case 1: 2

Case 2: 11

Note

Dataset is huge. Use faster i/o methods.

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PROBLEM SETTER: MOHIUL ALAM PRINCE

SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET)

题目大意：求是m的倍数并且连续的子序列的个数

解析：根据前缀和求

代码如下：

#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<set>#include<string>#include<cstdio>#include<cstring>#include<cctype>#include<cmath>#define N 100009using namespace std;const int inf = 0x3f3f3f3f;const int mod = 1e9 + 7;const double eps = 1e-8;const double pi = acos(-1.0);typedef long long LL;int a[N], s[N];int main(){    int t, cnt = 0;    int i, num;    cin >> t;    while(t--)    {        memset(s, 0, sizeof(s));        int n, m;        scanf("%d%d", &n, &m); a[0] = 0;        for(i = 1; i <= n; i++)        {            scanf("%d", &num);            a[i] = (a[i - 1] + num) % m;            s[a[i]]++;        }        LL ans = 0;        for(i = 0; i < m; i++)            ans += (LL)s[i] * ((LL)s[i] - 1LL) / 2LL;        printf("Case %d: %lld\n", ++cnt, ans + s[0]);    }    return 0;}