lightoj 1134 - Be Efficient

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

n代表输入4个数，问你有多少个子串和能被m整除。

先说一个定理：两个前缀和对m的余数相同，那么中间那段的和就是m的倍数。

dp[i]表示前缀和余数为i出现的次数。因为子串可以和前面的进行组合，所以出现2次的时候加1，出现3次的时候加2，以此类推。。。。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define inf 1e9#define endl '\n'#define LL long longint main(void){    int T,n,m,i,x;    int dp[100010];    scanf("%d",&T);    int cas = 1;    while(T--)    {        scanf("%d%d",&n,&m);        memset(dp,0,sizeof(dp));        int mod = 0;        LL ans = 0;        dp[0] = 1;        for(i=0;i<n;i++)        {            scanf("%d",&x);            mod = (mod + x)%m;            ans += dp[mod];            dp[mod]++;        }        printf("Case %d: %lld\n",cas++,ans);    }}