CodeForces 401D 数位DP

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=98182#problem/D

Description

Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.

Number x is considered close to number n modulo m, if:

• it can be obtained by rearranging the digits of number n,
• it doesn't have any leading zeroes,
• the remainder after dividing number x by m equals 0.

Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.

Input

The first line contains two integers: n(1 ≤ n < 10¹⁸) and m(1 ≤ m ≤ 100).

Output

In a single line print a single integer — the number of numbers close to number n modulo m.

Sample Input

Input

104 2

Output

3

Input

223 4

Output

1

Input

7067678 8

Output

47

Hint

In the first sample the required numbers are: 104, 140, 410.

In the second sample the required number is 232.

/**CodeForces 401D  数位DP题目大意：给定一个数字n，将其所有的位上的数重组后组成新的数，问重组后的数有多少个是m的倍数解题思路：一开始想着统计0~9十个数各出现了多少个，然后普通dp的，费了半天劲，后来觉得状态太大无法表示，才想到用数位dp。dp[s][k]表示每个数位上的数使用的状态为s（5二进制表示101，那么就是第一三位上的数已使用，第二位上的数没有使用）          当前数对m的取模为k的数有多少个。有一个问题就是如果直接dp会重复取的，比如：111 3 答案是1而不是6，那么用一个          flag数组表示第i位上bit[i]已用就不会出现重复取的情况了，然后dp一下就可以了。数位dp我用的kuangbin的模板*/#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;typedef long long LL;LL dp[(1<<19)+5][111],n;int m,bit[25],len;LL dfs(int pos,int s,int k){    if(pos<0&&(s==(1<<len)-1))return (k==0);    if(dp[s][k]!=-1)return dp[s][k];    LL ans=0;    int flag[10];    memset(flag,0,sizeof(flag));    for(int i=0;i<len;i++)    {        if((s&(1<<i))==0&&flag[bit[i]]==0)        {            flag[bit[i]]=1;            ans+=dfs(pos-1,s|(1<<i),(k*10+bit[i])%m);        }    }    dp[s][k]=ans;    return ans;}LL solve(LL n){    int k=0;    while(n)    {        bit[k++]=n%10;        n/=10;    }    len=k;    int s=0;    LL ans=0;    int flag[10];    memset(flag,0,sizeof(flag));    for(int i=0;i<len;i++)    {        if(bit[i]&&flag[bit[i]]==0)        {             flag[bit[i]]=1;            ans+=dfs(k-2,s|(1<<i),bit[i]%m);        }    }    return ans;}int main(){    while(~scanf("%I64d%d",&n,&m))    {        memset(dp,-1,sizeof(dp));        printf("%I64d\n",solve(n));    }    return 0;}/**123 3104 2401 27067678 8*/