CodeforcesRound#235(Div.2)D.RomanandNumbers(状压dp)_html/css

time limit per test

memory limit per test

input

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:

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?

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

input

223 4

output

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..

#include #include 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<