ZOJ_1241_Geometry Made Simple

本文介绍了一个简单的几何计算问题——已知直角三角形的两边求第三边的长度。通过给出示例输入输出及示例程序,展示了如何利用计算机快速解决这类数学问题。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Geometry Made Simple

Time Limit: 2 Seconds      Memory Limit: 65536 KB

Mathematics can be so easy when you have a computer. Consider the following example. You probably know that in a right-angled triangle, the length of the three sides a, b, c (where c is the longest side, called the hypotenuse) satisfy the relation a*a+b*b=c*c. This is called Pythagora's Law.

Here we consider the problem of computing the length of the third side, if two are given.

Input

The input contains the descriptions of several triangles. Each description consists of a line containing three integers a, b and c, giving the lengths of the respective sides of a right-angled triangle. Exactly one of the three numbers is equal to -1 (the 'unknown' side), the others are positive (the 'given' sides).

A description having a=b=c=0 terminates the input.


Output

For each triangle description in the input, first output the number of the triangle, as shown in the sample output. Then print "Impossible." if there is no right-angled triangle, that has the 'given' side lengths. Otherwise output the length of the 'unknown' side in the format "s = l", where s is the name of the unknown side (a, b or c), and l is its length. l must be printed exact to three digits to the right of the decimal point.

Print a blank line after each test case.


Sample Input

3 4 -1
-1 2 7
5 -1 3
0 0 0


Sample Output

Triangle #1
c = 5.000

Triangle #2
a = 6.708

Triangle #3
Impossible.


题意和思路:这个题目确实是水题。。没人做不出来吧?真心不想说思路了。。自己看代码吧。。


Sample Program Here

/******************************************
**
**      Author: Wan KaiMing
**      Date:   2012-11-02-20.07 星期五
**
*******************************************/


#include<iostream>
#include<string>
#include<cmath>
#include<iomanip>
using namespace std;
int main(){
   double a=0,b=0,c=0; //表示三角形的三条边a,b,c
   int count=0; //用于计数
   while(cin>>a>>b>>c){

      if(a==b&&b==c&&a==0) break; //全为0的时候退出
      cout<<"Triangle #"<<++count<<endl;

      if(a==-1){
        a=sqrt(c*c-b*b);
        if((c*c-b*b)<0) cout<<"Impossible."<<endl;
        else cout<<fixed<<setprecision(3)<<"a = "<<a<<endl;
      }
      else if(b==-1){
        b=sqrt(c*c-a*a);
        if((c*c-a*a)<0) cout<<"Impossible."<<endl;
        else cout<<fixed<<setprecision(3)<<"b = "<<b<<endl;
      }
      else{
        c=sqrt(a*a+b*b);
        cout<<fixed<<setprecision(3)<<"c = "<<c<<endl;
      }

      cout<<endl;
   }

   return 0;
}



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值