zoukankan      html  css  js  c++  java
  • hdu-5858 Hard problem(数学)

    题目链接:

    Hard problem

    Time Limit: 2000/1000 MS (Java/Others)   

     Memory Limit: 65536/65536 K (Java/Others)


    Problem Description
    cjj is fun with math problem. One day he found a Olympic Mathematics problem for primary school students. It is too difficult for cjj. Can you solve it?


    Give you the side length of the square L, you need to calculate the shaded area in the picture.

    The full circle is the inscribed circle of the square, and the center of two quarter circle is the vertex of square, and its radius is the length of the square.
     
    Input
    The first line contains a integer T(1<=T<=10000), means the number of the test case. Each case contains one line with integer l(1<=l<=10000).
     
    Output
    For each test case, print one line, the shade area in the picture. The answer is round to two digit.
     
    Sample Input
    1
    1
     
    Sample Output
    0.29
     
    题意:
     
    求变成为l的正方形中阴影部分的面积;
     
    思路:
     
    建坐标系积分就好了;以正方形的中心为原点,以对角线为水平轴,然后得到两个圆的方程,联立求交点得到积分区间,然后积分,积分的时候借用三角函数;
     
    AC代码:
     
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <bits/stdc++.h>
    #include <stack>
    #include <map>
     
    using namespace std;
     
    #define For(i,j,n) for(int i=j;i<=n;i++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
     
    typedef  long long LL;
     
    template<class T> void read(T&num) {
        char CH; bool F=false;
        for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
        for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
        F && (num=-num);
    }
    int stk[70], tp;
    template<class T> inline void print(T p) {
        if(!p) { puts("0"); return; }
        while(p) stk[++ tp] = p%10, p/=10;
        while(tp) putchar(stk[tp--] + '0');
        putchar('
    ');
    }
     
    const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const int inf=1e9;
    const int N=5e5+10;
    const int maxn=(1<<20)+14;
    const double eps=1e-12;
    
    
    double s;
    double solve(double x)
    {
        return (2*x+sin(2*x))/4;
    }
    inline void Init()
    {
         s=sqrt(7)/2;
         //cout<<s<<endl;
         s+=solve(asin(sqrt(14)/4))-4*solve(asin(sqrt(14)/8));
    }
    int main()
    {
        int t;
        read(t);
        Init();
        while(t--)
        {
            double l;
            scanf("%lf",&l);
            printf("%.2lf
    ",s*l*l);
        }
        return 0;
    }
    

      

  • 相关阅读:
    Oracle查看锁表并杀死进程
    Oracle查看表空间、是否开启扩展
    Linux日常命令记录
    【转载】LUbuntu,Ubuntu下设置自定义桌面分辨率
    windows下查找端口并杀死进程
    【转载】JAVA中IO流操作的基本规律总结
    【转载】hashCode()、equals()以及compareTo()方法的理解
    RuntimeException、Exception联系区别
    Java内存、数组
    【概念理解】同步异步、阻塞非阻塞
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5784688.html
Copyright © 2011-2022 走看看