Codeforces Round #235 (Div. 2) D. Roman and Numbers(状压dp)

Roman and Numbers

time limit per test
4 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

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 test(s)

input

104 2

output

3

input

223 4

output

1

input

7067678 8

output

47

Note

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

In the second sample the required number is 232.

题意：给出一个数字num和m，问通过重新排列num中的各位数字中有多少个数(mod m)=0,直接枚举全排列肯定不行，可以用状压dp来搞..

dp[S][k]表示选了num中的S且(mod m)=k的方案种数,初始条件dp[0][0]=1,转移方为dp[i|1<<j[(10*k+num[j])%m]+=dp[i}[k];,注意到是多重排列，所以还需要除去重复的。代码如下：

#include <iostream>
#include <cstring>
using namespace std;

typedef long long ll;

ll dp[1<<18][100],c[20];//dp[S][k]表示选了num中的S且(mod m)=k的方案种数

int main(int argc, char const *argv[])
{
char num[20];
int m;
while(cin>>num>>m) {

memset(dp,0,sizeof dp);
memset(c,0,sizeof c);
dp[0][0]=1;

ll div=1,sz=strlen(num),t=1<<sz;
for(int i=0;i<sz;i++) {
div*=(++c[num[i]-='0']);//可重排列最后要除的除数n1!*n2!*...nk!
}

for(int i=0;i<t;i++) {
for(int j=0;j<sz;j++)if(!(i&1<<j)) {//集合S中不含j才转移
if(num[j]||i){//至少一个不为0保证无前导0
for(int k=0;k<m;k++) {
dp[i|1<<j][(10*k+num[j])%m]+=dp[i][k];
}
}
}
}

cout<<dp[t-1][0]/div<<endl;
}
return 0;
}