codeforces #235 D. Roman and Numbers 题解

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【原题】

D. 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

input

223 4

output

input

7067678 8

output

Note

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

In the second sample the required number is 232.

【大意】给你一个n，让你找有多少x满足：x mod m=0。其中x是n的一个排列。重复的只计算一种。

【分析】这真是一道状态压缩的好题目。什么是状态压缩呢？就是把一个状态串压成一个十进制数。在这题里，对于n中的每一位数，都能表示成0或1（取或是不取）。而且总的状态数只有2^len-1（len是n的位数），因此我们可以用DP推出每一种状态。f[i][j]表示在i这种状态（01串的十进制形式），余数为j时的方案数。显然，如果我想取当前的数第K个数——dight[k]，f[i][j]就能推到f[i+2^(k-1)][(j*10+dight[k])%m]。当然，要注意这个位置是否已经用过，同时也要注意数字的判重情况。

【代码】

#include<cstdio>
#include<cstring>
using namespace std;
const int size=19;
bool flag[10];
long long dight[size],f[1<<size][101],i,j,p,m,k,n,t;
int main()
{scanf("%I64d%I64d",&p,&m);while (p){dight[++n]=p%10;p/=10;}for (i=1;i<=n/2;i++) t=dight[i],dight[i]=dight[n-i+1],dight[n-i+1]=t;f[0][0]=1;for (i=0;i<(1<<n)-1;i++)  //枚举每一种状态。而且可以证明，到某一种状态时，它一定在之前被推完了。for (j=0;j<m;j++){memset(flag,0,sizeof(flag));  //防止出现相同的情况。if (f[i][j]==0) continue;    //这种状态本身已经推不到。for (k=1;k<=n;k++)if (!flag[dight[k]])      //某一个数码没有出现过。{if (i&(1<<(k-1))) continue;  //这一位已经用过了if (i==0&&dight[k]==0) continue;  //前导0的问题flag[dight[k]]=true;f[i+(1<<(k-1))][(j*10+dight[k])%m]+=f[i][j];  //递推}}printf("%I64d",f[(1<<n)-1][0]);return 0;
}

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