zoukankan      html  css  js  c++  java
  • Codeforces Round #235 (Div. 2) D. Roman and Numbers(如压力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。问通过又一次排列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,能够用状压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。所以还须要除去反复的。

    代码例如以下:


    #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;
    }


    版权声明:本文博主原创文章。博客,未经同意不得转载。

  • 相关阅读:
    公司到底是怎么看我们的……[转]
    C# String.Format 的使用[转]
    Oracle9I 在安装时出现[登台区出现问题,请确保指定有效的“源”和“目标”!]
    街机游戏下载
    C#操作Excel时直接引用Com和InteropExcel的差异
    我所理解的接口和抽象类[转]
    c# winFrom 使窗体显示SplitContainer或Panel中[转]
    SQL Server 2005之PIVOT/UNPIVOT行列转换(转)
    Solaris大半年使用感触
    solaris上的pkg管理
  • 原文地址:https://www.cnblogs.com/mengfanrong/p/4854143.html
Copyright © 2011-2022 走看看