CodeforcesRound#235(Div.2)D.RomanandNumbers(狀壓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.

題意：給出一個數字num和m，問通過重新排列num中的各位數字中有多少個數(mod m)=0,直接枚舉全排列肯定不行，可以用狀壓dp來搞..

#include #include using namespace std;typedef long long ll;ll dp[1<<18][100],c[20];//dp[S][k]表示選了num中的S且(mod m)=k的方案種數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<