zoukankan      html  css  js  c++  java
  • ural One-two, One-two 2

     One-two, One-two 2
    Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
     
    Description
    A year ago the famous gangster Vito Maretti woke up in the morning and realized that he was bored of robbing banks of round sums. And for the last year he has been taking from banks sums that have only digits 1 and 2 in their decimal notation. After each robbery, Vito divides the money between N members of his gang. Your task is to determine the minimal stolen sum which is a multiple of N.

    Input

    The input contains the number N (1 ≤  N ≤ 10 6).

    Output

    Output the minimal number which is a multiple of N and whose decimal notation contains only digits 1 and 2. If it contains more than 30 digits or if there are no such numbers, then output "Impossible".

    Sample Input

    inputoutput
    5
    
    Impossible
    
    8
    
    112

    AC代码:

     1 #include<iostream>
     2 #include<cstring>
     3 #include<cstdio>
     4 #define N 5000002
     5 using namespace std;
     6 int n;         
     7 struct Node{
     8     int dig,mod;
     9     int f;//模拟指针,指向数组的下一个元素 
    10 }nod[N];
    11 
    12 int l,r;
    13 bool vis[N];
    14 int ret;
    15 
    16 void BFS(){
    17     while(l<r){
    18         int tmod1=(nod[l].mod*10+1)%n;
    19         int tmod2=(nod[l].mod*10+2)%n;
    20         if(!vis[tmod1]){
    21           vis[tmod1]=1;
    22             r++;
    23             nod[r].mod=tmod1;   
    24             nod[r].dig=1;
    25             nod[r].f=l;
    26             if(tmod1==0){ret=r;return ;}
    27         }
    28         if(!vis[tmod2]){
    29           vis[tmod2]=1;
    30             r++;
    31             nod[r].mod=tmod2;
    32             nod[r].dig=2;
    33             nod[r].f=l;
    34             if(tmod2==0){ret=r;return ;}
    35         }
    36 
    37         l++;
    38 
    39     }
    40 }
    41 
    42 void out(int x)
    43 {
    44     if(x>0)out(nod[x].f);
    45     else if(x==0)return;
    46     cout<<nod[x].dig;
    47 }
    48 
    49 void first()
    50 {
    51    nod[1].dig=1;
    52    nod[1].f=0;
    53    nod[1].mod=1;
    54    nod[2].dig=2;
    55    nod[2].f=0;
    56    nod[2].mod=2;
    57 }
    58 
    59 int main()
    60 {
    61     while(cin>>n)
    62     {
    63         if(n==1||n==2){
    64             cout<<n<<endl;
    65             continue;
    66         }
    67         first();
    68         
    69         l=1;r=2;ret=0;
    70         memset(vis,0,sizeof(vis));
    71         vis[1]=vis[2]=1;
    72        
    73         BFS();
    74         if(ret==0){
    75             cout<<"Impossible"<<endl;
    76             continue;
    77         }
    78         out(ret);
    79         cout<<endl;
    80     }
    81     return 0;
    82 }
  • 相关阅读:
    基于HTTP协议的轻量级简单队列服务-HTTPSQS
    PHP获取客户端IP
    编译安装Memcached并使用systemctl管理
    win10利用WSL2安装docker的2种方式
    如何提升前端基建的效能价值?
    如何衡量前端基建的效能价值?
    从面向对象角度看前端工程体系
    「前端工程化」该怎么理解?
    跨端方案的三大困境
    React 17 要来了,非常特别的一版
  • 原文地址:https://www.cnblogs.com/liugl7/p/4843423.html
Copyright © 2011-2022 走看看