题意:告诉你两个圆环,求圆环相交的面积. /* gyt Live up to every day */ #include<cstdio> #include<cmath> #include<iostream> #include<algorithm> #include<vector> #include<stack> #include<cstring> #include<queue> #include<set&…

题目链接: POJ:http://poj.org/problem? id=2546 ZOJ:problemId=597" target="_blank">http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=597 Description Your task is to write a program, which, given two circles, calculates the area of the…

已知两圆圆心坐标和半径,求相交部分面积: #include <iostream> using namespace std; #include<cmath> #include<stdio.h> #define PI 3.141593 struct point//点 { double x,y; }; struct circle//圆 { point center; double r; }; float dist(point a,point b)//求圆心距 { return…

题目传送门 Hard problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1066    Accepted Submission(s): 622 Problem Description cjj is fun with math problem. One day he found a Olympic Mathematics pr…

转载 两圆相交分如下集中情况:相离.相切.相交.包含. 设两圆圆心分别是O1和O2,半径分别是r1和r2,设d为两圆心距离.又因为两圆有大有小,我们设较小的圆是O1. 相离相切的面积为零,代码如下: double d = sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); if (d >= r1+r2) return 0; double d = sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); if (…

Intersection Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others) Problem Description Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examp…

题目链接 题意 : 给你两个圆的半径和圆心,让你求两个圆相交的面积大小. 思路 : 分三种情况讨论 假设半径小的圆为c1,半径大的圆为c2. c1的半径r1,圆心坐标(x1,y1).c2的半径r2,圆心坐标(x2,y2). d为两圆圆心连线的长度. 相交面积为S d=sqrt((x1-x2)^2+(y1-y2)^2) (1)如果r1+r2<=d 那么两圆相离,相交面积S=0 (2)如果r2-r1>=d 那么半径小的圆内含半径大的圆,那么相交面积为小圆的面积S=pi*r1*r1 (3)既非(1)…

Open-air shopping malls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2344    Accepted Submission(s): 866 Problem Description The city of M is a famous shopping city and its open-air shopping…

题目大意是:先给你一些圆,你可以任选这些圆中的一个圆点作圆,这个圆的要求是:你画完以后.这个圆要可以覆盖之前给出的每一个圆一半以上的面积,即覆盖1/2以上每一个圆的面积. 比如例子数据,选左边还是选右边没差别,红色的圆为答案(选了左边的圆点),它覆盖了左边圆的1/2以上,也覆盖了右边圆的1/2以上. 知道了怎样求两圆面积交.那么这道题就简单了.仅仅要二分答案,然后枚举每个圆点,假设全都覆盖了1/2以上就继续二分,最后答案就得出来了. #include<iostream> #include<…

链接 画图推公式 这两种情况 都可用一种公式算出来 就是两圆都求出圆心角 求出扇形的面积减掉三角形面积 #include <iostream> using namespace std; #include<cmath> #include<iomanip> #include<algorithm> int main() { double d,t,t1,s,x,y,xx,yy,r,rr; while(cin>>x>>y>>r) {…