zoukankan      html  css  js  c++  java
  • hdu2393Higher Math

    Problem Description
    You are building a house. You’d prefer if all the walls have a precise right angle relative to the ground, but you have no device to measure angles. A friend says he has a great idea how you could ensure that all walls are upright: All you need to do is step away a few feet from the wall, measure how far away you are from the wall, measure the height of the wall, and the distance from the upper edge of the wall to where you stand. You friend tells you to do these measurements for all walls, then he’ll tell you how to proceed. Sadly, just as you are done, a timber falls on your friend, and an ambulance brings him to the hospital. This is too bad, because now you have to figure out what to do with your measurements yourself.

    Given the three sides of a triangle, determine if the triangle is a right triangle, i.e. if one of the triangle’s angles is 90 degrees.
     
    Input
    The inputs start with a line containing a single integer n. Each of the n following lines contains one test case. Each test case consists of three integers 1 <= a, b, c <= 40000 separated by a space. The three integers are the lengths of the sides of a triangle.
     
    Output
    The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario counting from 1. After that, output a single line containing either the string “yes” or the string “no”, depending on if the triangle in this test case has a right angle. Terminate each test case with an empty line.
     
    Sample Input
    2 36 77 85 40 55 69
     
    Sample Output
    Scenario #1: yes Scenario #2: no
     
    Source
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<cmath>
    
    using namespace std;
    
    int main()
    {
        int z,t = 0;
        cin>>z;
        while(z--)
        {
            printf("Scenario #%d:
    ",++t);
            int a,b,c;
            cin>>a>>b>>c;
            if(a>b) swap(a,b);
            if(b>c) swap(b,c);
            if(a*a+b*b == c*c) cout<<"yes"<<endl<<endl;
            else cout<<"no"<<endl<<endl;
        }
        return 0;
    }
  • 相关阅读:
    1061 Dating (20 分)
    1042 Shuffling Machine (20 分)简单模拟
    1132 Cut Integer (20 分)
    1100 Mars Numbers (20 分)
    1077 Kuchiguse (20 分)求字符串最长相同后缀
    1065 A+B and C (64bit) (20 分)大数 溢出
    1107 Social Clusters (30 分)并查集
    1079 Total Sales of Supply Chain (25 分)
    1078 Hashing (25 分)
    1063 Set Similarity (25 分)
  • 原文地址:https://www.cnblogs.com/wos1239/p/4557070.html
Copyright © 2011-2022 走看看