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; }
 

CF401D Roman and Numbers 状压DP的更多相关文章

  1. Codeforces Round #235 (Div. 2) D. Roman and Numbers 状压dp+数位dp

    题目链接: http://codeforces.com/problemset/problem/401/D D. Roman and Numbers time limit per test4 secon ...

  2. CF449D Jzzhu and Numbers (状压DP+容斥)

    题目大意: 给出一个长度为n的序列,构造出一个序列使得它们的位与和为0,求方案数 也就是从序列里面选出一个非空子集使这些数按位与起来为0. 看了好久才明白题解在干嘛,我们先要表示出两两组合位与和为0的 ...

  3. 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 i ...

  4. [poj2411] Mondriaan's Dream (状压DP)

    状压DP Description Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One nigh ...

  5. Codeforces Round #321 (Div. 2) D. Kefa and Dishes 状压dp

    题目链接: 题目 D. Kefa and Dishes time limit per test:2 seconds memory limit per test:256 megabytes 问题描述 W ...

  6. Codeforces Gym 100610 Problem K. Kitchen Robot 状压DP

    Problem K. Kitchen Robot Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10061 ...

  7. TZOJ 2289 Help Bob(状压DP)

    描述 Bob loves Pizza but is always out of money. One day he reads in the newspapers that his favorite ...

  8. Educational Codeforces Round 13 E. Another Sith Tournament 状压dp

    E. Another Sith Tournament 题目连接: http://www.codeforces.com/contest/678/problem/E Description The rul ...

  9. Codeforces Beta Round #8 C. Looking for Order 状压dp

    题目链接: http://codeforces.com/problemset/problem/8/C C. Looking for Order time limit per test:4 second ...

随机推荐

  1. style属性css与javascript对照表

    有时候会用javascript来控制标签的style,但js的style属性写法跟css有点不一样,通常是一个单词的写法不变,单词-单词属性会去掉“-”,再把第二个单词的首字母大写,估计是为了与减法运 ...

  2. 关于ArrayList源码

    一.构造方法 private static final int DEFAULT_CAPACITY = 10; //空参的构造方法,初始化数组长度为默认值,默认值为10 public ArrayList ...

  3. HttpClient远程接口调用-实名认证

    1.HttpClient远程接口调用 1)用户注册 注册按钮button提交表单时,要return false form表单 <!-- action="http://localhost ...

  4. [Leetcode] 第331题 验证二叉树的前序序列化

    一.题目描述 序列化二叉树的一种方法是使用前序遍历.当我们遇到一个非空节点时,我们可以记录下这个节点的值.如果它是一个空节点,我们可以使用一个标记值记录,例如 #. _9_ / \ 3 2 / \ / ...

  5. java自学小段 产生随机数

    public class Suijishu { public static void main(String[] args) { double i=Math.random();//产生一个0-0.5的 ...

  6. 决策树(基于增益率)之python实现

    如图,为使用到的公式,信息熵表明样本的混乱程度,增益表示熵减少了,即样本开始分类,增益率是为了平衡增益准则对可取值较多的属性的偏好,同时增益率带来了对可取值偏小的属性的偏好,实际中,先用增益进行筛选, ...

  7. html标签和css基础语法与浏览器兼容性等相关基础学习

    <!-- table的使用 --> <h3>前端日常</h3> <form action="https://www.baidu.com"& ...

  8. jedis 2.9版本部分属性变更

    1.控制一个pool可分配多少个jedis实例 “maxActive” -> “maxTotal” 2.最大建立连接等待时间.如果超过此时间将接到异常.设为-1表示无限制. “maxWait” ...

  9. jQuery常用方法(三)-jQuery Ajax

    JQuery Ajax 方法说明: load( url, [data], [callback] ) 装入一个远程HTML内容到一个DOM结点. $("#feeds").load(& ...

  10. Python celery和Redis入门安装使用(排难帖)

    1.redis安装 下载地址 https://github.com/MicrosoftArchive/redis/releases,选择Redis-x64-3.2.100.msi5.8 MB下载就好了 ...