zoukankan      html  css  js  c++  java
  • Coin Test

    描述

    As is known to all,if you throw a coin up and let it droped on the desk there are usually three results. Yes,just believe what I say ~it can be the right side or the other side or standing on the desk, If you don't believe this,just try In the past there were some famous mathematicians working on this .They repeat the throwing job once again. But jacmy is a lazy boy.He is busy with dating or playing games.He have no time to throw a single coin for 100000 times. Here comes his idea,He just go bank and exchange thousands of dollars into coins and then throw then on the desk only once. The only job left for him is to count the number of coins with three conditions.

    He will show you the coins on the desk to you one by one. Please tell him the possiblility of the coin on the right side as a fractional number if the possiblity between the result and 0.5 is no larger than 0.003. BE CAREFUL that even 1/2,50/100,33/66 are equal only 1/2 is accepted ! if the difference between the result and 0.5 is larger than 0.003,Please tell him "Fail".Or if you see one coin standing on the desk,just say "Bingo" any way.

    输入
    Three will be two line as input.
    The first line is a number N(1<N<65536)
    telling you the number of coins on the desk.
    The second line is the result with N litters.The letter are "U","D",or "S","U" means the coin is on the right side. "D" means the coin is on the other side ."S" means standing on the desk.
    输出
    If test successeded,just output the possibility of the coin on the right side.If the test failed please output "Fail",If there is one or more"S",please output "Bingo"
    样例输入
    6
    UUUDDD
    样例输出
    1/2

     1 #include <stdio.h>
     2 #include <math.h>
     3 
     4 int gcd(int a,int b);
     5 
     6 int main(){
     7     int n;
     8     int i;
     9     char c;
    10     int U_number;
    11     int S_number;
    12     double result;
    13     
    14     scanf("%d",&n);
    15     getchar();
    16     
    17     U_number=0;
    18     S_number=0;
    19     for(i=1;i<=n;i++){
    20         c=getchar();
    21         
    22         if(c=='U')
    23             U_number++;
    24             
    25         else if(c=='S')
    26             S_number++;
    27     }
    28     
    29     result=(double)U_number/n;
    30     if(fabs(result-0.5)<=0.003)    
    31         printf("%d/%d
    ",U_number/gcd(U_number,n),n/gcd(U_number,n));
    32     
    33     else{
    34         if(S_number>=1)
    35             printf("Bingo
    ");
    36             
    37         else
    38             printf("Fail
    ");
    39     }
    40     return 0;
    41 }
    42 
    43 int gcd(int a,int b){
    44     int temp;
    45     
    46     if(a<b){
    47         temp=a;
    48         a=b;
    49         b=temp;
    50     }
    51     
    52     while(a%b!=0){
    53         temp=a%b;
    54         a=b;
    55         b=temp;
    56     }
    57     
    58     return b;
    59 }
  • 相关阅读:
    Two Sum 2015年6月8日
    Word Ladder II 2015年6月4日
    Word Ladder 2015年6月3日
    华为
    《安卓网络编程》之第八篇 安卓与服务器之间通讯JSON
    星辉信息科技Odoo开发教程10-odoo开发环境准备
    星辉信息科技Odoo开发教程9-odoo创建视图02
    星辉信息科技Odoo开发教程8-odoo创建视图01
    星辉信息科技Odoo开发教程7-创建菜单项
    星辉信息科技Odoo开发教程6-配置安全权限控制02
  • 原文地址:https://www.cnblogs.com/zqxLonely/p/4099207.html
Copyright © 2011-2022 走看看