链接:https://www.nowcoder.com/acm/contest/141/J来源:牛客网 Eddy has graduated from college. Currently, he is finding his future job and a place to live. Since Eddy is currently living in Tien-long country, he wants to choose a place inside Tien-long country
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
[题解]CIRU - The area of the union of circles [SP8073] \ 圆的面积并 [Bzoj2178] 传送门: \(\text{CIRU - The area of the union of circles [SP8073]}\) 圆的面积并 \(\text{[Bzoj2178]}\) [题目描述] 给出 \(n\) 个圆的圆心坐标 \((x,y)\) 和半径 \(r\),求它们覆盖的总面积. [输入] 第一行一个整数 \(n\),表示一共有 \(n\)
map函数 :处理序列中的每个元素,得到的结果是一个列表,该列表元素个数及位置与原来一样 ## 求列表里元素的平方 (原始方法) num_1=[1,2,13,5,8,9] res =[] for i in num_1: res.append(i**2) print('打印结果:',res) 打印结果: [1, 4, 169, 25, 64, 81] 有多个列表求里面元素的平方 (定义成函数) num_1=[1,2,13,5,8,9] def test(array): res =[] for i
[SPOJ-CIRU]The area of the union of circles/[BZOJ2178]圆的面积并 题目大意: 求\(n(n\le1000)\)个圆的面积并. 思路: 对于一个\(x\),我们可以用线段覆盖的方法求出被圆覆盖的长度.用\(f(x)\)表示横坐标为\(x\)时覆盖的长度,则我们可以对\(f(x)\)积分来得到答案.注意面积不连续的部分要分开求. 源代码: #include<cmath> #include<cstdio> #include<cc