zoukankan      html  css  js  c++  java
  • HOJ———丢手绢

    hide handkerchief

    Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 6646 Accepted Submission(s): 2173
     
    Problem Description
    The Children’s Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends.
    Now I introduce the game to you. Suppose there are N people played the game ,who sit on the ground forming a circle ,everyone owns a box behind them .Also there is a beautiful handkerchief hid in a box which is one of the boxes .
    Then Haha(a friend of mine) is called to find the handkerchief. But he has a strange habit. Each time he will search the next box which is separated by M-1 boxes from the current box. For example, there are three boxes named A,B,C, and now Haha is at place of A. now he decide the M if equal to 2, so he will search A first, then he will search the C box, for C is separated by 2-1 = 1 box B from the current box A . Then he will search the box B ,then he will search the box A.
    So after three times he establishes that he can find the beautiful handkerchief. Now I will give you N and M, can you tell me that Haha is able to find the handkerchief or not. If he can, you should tell me "YES", else tell me "POOR Haha".
     
    Input
    There will be several test cases; each case input contains two integers N and M, which satisfy the relationship: 1<=M<=100000000 and 3<=N<=100000000. When N=-1 and M=-1 means the end of input case, and you should not process the data.
     
    Output
    For each input case, you should only the result that Haha can find the handkerchief or not.
     
    Sample Input
    3 2
    -1 -1
     
    Sample Output
    YES
     

    最一开始并没有发现这道题的规律,只是盲目地将前几种情况列了出来。

    发现了一些“端倪”。。。然并卵,还是看了别人的博客

    http://blog.sina.com.cn/s/blog_86a6befd01012bqi.html

    发现了解法应该是利用辗转相除求解N和M-1的公约数

     1 #include <iostream>
     2 using namespace std;
     3 int main(){
     4     int n,m,t;
     5     while (1) {
     6         scanf("%d%d",&n,&m);
     7         if(n==-1&&m==-1)break;
     8         while (m!=0) {
     9             t=n%m;
    10             n=m;
    11             m=t;
    12         }
    13         if(n==1)printf("YES
    ");
    14         else printf("POOR Haha
    ");
    15     }
    16     return 0;
    17 }

    总结:1.无限循环tips-->while(1){}

  • 相关阅读:
    变参宏 __VA_ARGS__
    预处理中的 # 和 ##
    strlen与sizeof异同
    .vimrc
    sudo:有效用户 ID 不是 0,sudo 属于 root 并设置了 setuid 位吗
    远程ssh登陆时报错:/bin/bash: Permission denied
    Excel中VBA进行插入列、格式化、排序
    ORACLE发送带附件邮件的二三事之一
    Windows Server 2008 双网卡同时上内外网 不能正常使用
    VMWARE修改CPUID
  • 原文地址:https://www.cnblogs.com/MICE1024/p/5737253.html
Copyright © 2011-2022 走看看