zoukankan      html  css  js  c++  java
  • Star

    C. Star

    1000ms
    1000ms
    32768KB
     
     
    Submit Status 
    Font Size:  
    Overpower often go to the playground with classmates. They play and chat on the playground. One day, there are a lot of stars in the sky. Suddenly, one of Overpower’s classmates ask him: “How many acute triangles whose inner angles are less than 90 degrees (regarding stars as points) can be found? Assuming all the stars are in the same plane”. Please help him to solve this problem.
     

    Input

    The first line of the input contains an integer T (T≤10), indicating the number of test cases.

    For each test case:

    The first line contains one integer n (1≤n≤100), the number of stars.

    The next n lines each contains two integers x and y (0≤|x|, |y|≤1,000,000) indicate the points, all the points are distinct.

     

    Output

    For each test case, output an integer indicating the total number of different acute triangles.
     

    Sample Input

    1
    3
    0 0
    10 0
    5 1000
    
     

    Sample Output

    1
    //题意:每个样例给出n,然后n个坐标,求这些点能组成多少个锐角三角形
    //思路:只要知道锐角三角形三边的关系a^2+b^2>c^2即可.
    
    #include<stdio.h>
    #include<math.h>
    int main()
    {
        int t,n,i,j,k,time;
        double s[101][2];
        double a,b,c,x1,y1,x2,y2,x3,y3;
        scanf("%d",&t);
        while(t--)
        {
            time=0;
            scanf("%d",&n);
          for(i=0;i<n;i++)
              scanf("%lf%lf",&s[i][0],&s[i][1]);
           for(i=0;i<n;i++)
          {
              for(j=i+1;j<n;j++)
              {
                  for(k=j+1;k<n;k++)
                  {
                      x1=s[i][0];y1=s[i][1];
                      x2=s[j][0];y2=s[j][1];
                      x3=s[k][0];y3=s[k][1];
                     a=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
                     b=sqrt((x3-x1)*(x3-x1)+(y3-y1)*(y3-y1));
                     c=sqrt((x3-x2)*(x3-x2)+(y3-y2)*(y3-y2));
                     if(a*a+b*b>c*c&&a*a+c*c>b*b&&b*b+c*c>a*a)
                         time++;
                  }
              }
          }
          printf("%d
    ",time);
        }
        return 0;
    }
    //问题如果不加sqrt会超时,这是什么原因。。加了就ac了
  • 相关阅读:
    opencv4显示与保存图片
    opencv播放视频
    opencv4.1.0环境配置
    lambda表达式
    基于范围的for循环
    可调用对象包装器std::function
    C++11的类型推导
    Datagridview 实现二维表头
    Linux内存相关sysfs、工具
    关于net core 站点通过iis部署,跨域配置遇到的问题
  • 原文地址:https://www.cnblogs.com/cancangood/p/3252556.html
Copyright © 2011-2022 走看看