Light oj 1158 - Anagram Division（状压+记忆化）

1158 - Anagram Division

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

Given a string s and a positive integer d you have to determine how many permutations of s are divisible by d.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains a string s (1 ≤ s_length ≤ 10) and an integer d (1 ≤ d ≤ 1001). s will only contain decimal digits.

Output

For each case, print the case number and the number of permutations of s that are divisible by d.

Sample Input

Output for Sample Input

3

000 1

1234567890 1

123434 2

Case 1: 1

Case 2: 3628800

Case 3: 90

 

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PROBLEM SETTER: ABDULLAH AL MAHMUD

SPECIAL THANKS: JANE ALAM JAN (SOLUTION, DATASET)

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define bug printf("hihi\n")#define eps 1e-8typedef long long ll;using namespace std;#define INF 0x3f3f3f3f#define N 11int p[11];char c[N];int dp[1<<N][1005];int n,mod;int vis[N];void inint(){    p[0]=1;    for(int i=1;i<N;i++)        p[i]=p[i-1]*i;}int dfs(int cur,int le){   // printf("%d\n",cur);     if(cur==(1<<n)-1) return le ? 0:1;     if(cur>=(1<<n)) return 0;     if(dp[cur][le]!=-1) return dp[cur][le];     dp[cur][le]=0;     for(int i=0;i<n;i++)     {         if(cur&(1<<i)) continue;         dp[cur][le]+=dfs(cur|(1<<i),(le*10+c[i]-'0')%mod);     }    return dp[cur][le];}int main(){    inint();    int i,j,t,ca=0;    scanf("%d",&t);    while(t--)    {        scanf("%s%d",c,&mod);        n=strlen(c);        memset(vis,0,sizeof(vis));        int te=1;        for(int i=0;i<n;i++)            vis[c[i]-'0']++;        for(i=0;i<=9;i++)            te=te*p[vis[i]];        memset(dp,-1,sizeof(dp));        printf("Case %d: %d\n",++ca,dfs(0,0)/te);    }  return 0;}