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

1158 - Anagram Division

PDF (English)

Statistics

Forum

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

 

---

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-8
typedef long long ll;

using namespace std;

#define INF 0x3f3f3f3f
#define N 11

int 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;
}