zoukankan      html  css  js  c++  java
  • ZOJ 1241 Geometry Made Simple

    /*
    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.

    */

    #include<stdio.h>
    #include<math.h>
    
    int main(){
    	double a,b,c;
    	int n=0;
    	while(scanf("%lf%lf%lf",&a,&b,&c)){
    		if(a==0&&b==0&&c==0)
    			break;
    		n++;
    			if (c==-1)
                  {
                         c=sqrt(a*a+b*b);
                         printf("Triangle #%d
    ",n);
                         printf("c = %.3lf
    ",c);
                  }
                else
                    if (a==-1)
                   {
                         a=sqrt(c*c-b*b);
                         printf("Triangle #%d
    ",n);
                          if (c>b)
                         printf("a = %.3lf
    ",a);
                         else
                         printf("Impossible.
    ");
                   }
                     else
                     if (b==-1)
                    {
                       b=sqrt(c*c-a*a);
                       printf("Triangle #%d
    ",n);
                       if (c>a)
                       printf("b = %.3lf
    ",b);
                       else
                         printf("Impossible.
    ");
                    }
                    printf("
    ");
    
    	}
    
    	return 0;
    }
  • 相关阅读:
    无题
    生活真像小说
    草样年华
    Intentional Programming
    厌倦说话
    解释 Intentional Programming
    开始折腾iphone cdma 恢复,降级,刷机,越狱,手编,写号
    新手如何成为更好的图形设计师
    Grunt.js 初使用
    有抱负的程序员应看的10个TED演讲
  • 原文地址:https://www.cnblogs.com/naive/p/3568818.html
Copyright © 2011-2022 走看看