LightOJ

 Painful Bases

 LightOJ - 1021 

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

3

2 2

10

10 2

5681

16 1

ABCDEF0123456789

Sample Output

Case 1: 1

Case 2: 12

Case 3: 20922789888000

思路：状压DP。用d[i][k]表示在i这个状态，余数为k的排列有多少种。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;long long d[1<<16][21];char s[20];int base,mod;long long getsum(int i,int j){int th=0;while(i)th+=i%2,i/=2;long long sum=1;for(i=0;i<th;i++){sum*=base;sum%=mod;}if(s[j]>='0'&&s[j]<='9')return (sum*(s[j]-'0'))%mod;return (sum*(s[j]-'A'+10))%mod;}int main(){int T,n,cas=1;cin>>T;while(T--){cin>>base>>mod;scanf("%s",s);n=strlen(s);memset(d,0,sizeof d);d[0][0]=1;for(int i=0;i<(1<<n);i++){for(int j=0;j<n;j++){if(i&(1<<j))continue;long long sum=getsum(i,j); //求出j加到当前状态的值for(int k=0;k<mod;k++)d[i|(1<<j)][(k+sum)%mod]+=d[i][k];}}printf("Case %d: %lld\n",cas++,d[(1<<n)-1][0]);}return 0;}