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Time Limit: 2 second(s)

Memory Limit: 32 MB

As you know that sometimes base conversion is a painful task. But still there are interesting facts in bases.

For convenience let's assume that we are dealing with the bases from 2 to 16. The valid symbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. And you can assume that all the numbers given in this problem are valid. For example 67AB is not a valid number of base 11, since the allowed digits for base 11 are 0 to A.

Now in this problem you are given a base, an integer K and a valid number in the base which contains distinct digits. You have to find the number of permutations of the given number which are divisible by K. K is given in decimal.

For this problem, you can assume that numbers with leading zeroes are allowed. So, 096 is a valid integer.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a blank line. After that there will be two integers, base (2 ≤ base ≤ 16) and K (1 ≤ K ≤ 20). The next line contains a valid integer in that base which contains distinct digits, that means in that number no digit occurs more than once.

Output

For each case, print the case number and the desired result.

Sample Input	Output for Sample Input
3   2 2 10   10 2 5681   16 1 ABCDEF0123456789	Case 1: 1 Case 2: 12 Case 3: 20922789888000

 

題目連結：http://lightoj.com/volume_showproblem.php?problem=1021

題目大意：base表示base進位制，給一個k(0 <= k <= 20)，給一個base進位制下的合法數，保證每位數字都不同，求這個數字的所有排列中是k的倍數的個數

題目分析：因為數字都不同，所以最多隻有16位，因此可以狀壓做，dp[i][j]表示選數狀態為i，模k為j的排列數個數，然後對一個合法狀態擴充套件就是在前面加數字，取模的時候因為((x * base) + num[idx]) % k == ((x * base) % k + num[idx] % k) % k == (j * base + num[idx] % k ) % k所以轉移方程就是

dp[i | (1 << idx)][(j * base + num[idx] % k) % k] += dp[i][j]

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = (1 << 16) + 5;
ll dp[MAX][25];
char s[25];
int num[25];

int main()
{
    int T;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++)
    {
        int b, k;
        scanf("%d %d", &b, &k);
        scanf("%s", s);
        int len = strlen(s);
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < len; i++)
            num[i] = s[i] >= 'A' ? (10 + s[i] - 'A') : (s[i] - '0');
        dp[0][0] = 1;
        for(int i = 0; i < (1 << len); i++)
            for(int j = 0; j < k; j++)
                if(dp[i][j])
                    for(int idx = 0; idx < len; idx++)
                        if(!(i & (1 << idx)))
                            dp[i | (1 << idx)][(j * b + num[idx] % k) % k] += dp[i][j];
        printf("Case %d: %lld\n", ca, dp[(1 << len) - 1][0]);
    }
}