CF401D Roman and Numbers 状压DP

CF401D

题意翻译

将n(n<=10^18)的各位数字重新排列（不允许有前导零） 求 可以构造几个mod m等于0的数字

题目描述

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 n n , modulo m m m .

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

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

输入输出格式

输入格式：

The first line contains two integers: n n n $ (1<=n&lt;10^{18}) $ and m m m (1<=m<=100) (1<=m<=100) (1<=m<=100) .

输出格式：

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

输入输出样例

输入样例#1： 

104 2

输出样例#1： 

3

输入样例#2： 

223 4

输出样例#2： 

1

输入样例#3： 

7067678 8

输出样例#3： 

47

说明

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

In the second sample the required number is 232.

分析：

题目描述确实比较吓人， n位数字重新排列最多可以创造出多少个%m == 0 的数；

其实就是状态压缩；

定义f [i] [j] ， i表示一个二进制数， 1代表选这个数， j代表由n个数中选出x个组成的数%m==j；

 

转移方程 ： f[i|(1 << k)][(j * 10 + x) % m] += f[i][j]；

意义：对于第k位数x， 都可以由不选他转移到选他， 就是 i -> i |（1 << k）;

然后第二维就由 j -> (j *10 + x) % m   （显然）；

注意 ： 因为状态压缩是暴力的把每一位数当成与前边的数都不一样， 比如 11 ，应该算一次， 但是我们却算了两次；

方案 ： 1. 最后除以cnt！（cnt为一个数出现了多少次）。

　　　　2. 直接去重。

代码奉上：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
#define maxn (1 << 18) + 5
#define int long long

int n, m;

int s[21];

int f[maxn][105];

char ch[20];

bool vis[20];

signed main()
{
	scanf("%s%lld", &ch, &m);
	
	int n = strlen(ch);
	
	f[0][0] = 1;
	
	int e = (1 << n);
	for(register int i = 0 ; i < e ; i ++)
	{
		for(register int j = 0 ; j < m ; j ++)
		{
			memset(vis, 0, sizeof vis);
			for(register int k = 0 ; k < n ; k ++)
			{
				int x = ch[k] - '0';
				if(i & (1 << k)) continue;
				if(i == 0 && x == 0) continue;
				if(vis[x]) continue;
				vis[x] = 1;
				f[i|(1<<k)][(j*10+x)%m] += f[i][j];
			}
		}
	}
	
	cout << f[e-1][0];
	return 0;

	
}