Consider an integer sequence consisting of N elements, where -

X₀ = A

X_i = (( (X_i-1) * B + C ) % M)+1 ���� for i = 1 to N-1

You will be given the values of A, B, C, M and N. Find out the number of� �consecutive subsequences� whose sum is a multiple of M.

Consider an example where A = 2, B = 1, C = 2, M = 4 and N = 4.

So, X₀ = 2, X₁ = 1, X₂ = 4 and X₃ = 3.

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

The first line of input is an integer T(T<500) that indicates the number of test cases. Eact case consists of 5 integers A, B, C, M and N. A, B and C will be non-negative integers not greater than 1000.� N and M will be a positive integers not greater than 10000.

Output

For each case, output the case number followed by the result.

2

2 1 2 4 4

923 278 195 8685 793

Case 1: 2

Case 2: 34

Problem Setter: Mohiul Alam Prince
Moderator and Alternate Solution: Sohel Hafiz

題目描述：

根據給定的方程將陣列建出來，問有多少組區間總合是 M 的倍數。

題目解法：

記錄一個累計值 A[i] = sigma(X[i])%M。

對於一個區間 X[l ... r] 的總合，可以看成 A[r]-A[l-1]，如果存在是 M 的倍數，

那麼 A[l] 和 A[r] 具有相同的餘數，因此用掃描線的思路去看，找 0...i-1 有多少組餘數等於 A[i]。

但是如果餘數為 0 要特別考慮自己本身。

#include <stdio.h>
int X[10005];
int main() {
    int A, B, C, M, N;
    int testcase, cases = 0;
    int i;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d %d %d %d", &A, &B, &C, &M, &N);
        X[0] = A;
        for(i = 1; i < N; i++)
            X[i] = (X[i-1]*B+C)%M+1;
        int sum = 0, mark[10005] = {};
        int ret = 0;
        for(i = 0; i < N; i++) {
            sum = (sum+X[i])%M;
            ret += mark[sum];
            if(sum == 0)    ret++;
            mark[sum]++;
        }
        printf("Case %d: %d\n", ++cases, ret);
    }
    return 0;
}