HDU 1798 Tell me the area (数学)】的更多相关文章

题目链接 Problem Description     There are two circles in the plane (shown in the below picture), there is a common area between the two circles. The problem is easy that you just tell me the common area.   Input There are many cases. In each case, there…
http://acm.hdu.edu.cn/showproblem.php?pid=1798 Problem Description     There are two circles in the plane (shown in the below picture), there is a common area between the two circles. The problem is easy that you just tell me the common area.   Input…
题意:求阴影部分面积. 析:没什么可说的,就是一个普通的定积分. 代码如下: #include <cstdio> #include <iostream> using namespace std; int main(){ int T; cin >> T; double x0, y0, x1, y1, x2, y2, k, b, a, c, h, s; while(T--){ scanf("%lf %lf %lf %lf %lf %lf",&x0…
History repeat itself Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 434264-bit integer IO format: %I64d      Java class name: Main Prev Submit Status Statistics Discuss Next Type: None   None Graph Theory 2…
find the nth digit Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7403    Accepted Submission(s): 2120 Problem Description 假设:S1 = 1S2 = 12S3 = 123S4 = 1234.........S9 = 123456789S10 = 12345678…
2019 杭电多校 8 1003 题目链接:HDU 6659 比赛链接:2019 Multi-University Training Contest 8 Problem Description Acesrc is a famous mathematician at Nanjing University second to none. Playing with interesting numbers is his favorite. Today, he finds a manuscript whe…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5019 Problem Description In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), highest common factor (hcf), or greatest common measure (gcm), of two or more i…
Explore Track of Point Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5476 Description In Geometry, the problem of track is very interesting. Because in some cases, the track of point may be beautiful curve. For…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4091 /** 这题的一种思路就是枚举了: 基于这样一个事实:求出lcm = lcm(s1,s2), num1 = lcm/s1, num2 = lcm/s2; 则价值与体积比小的那个宝藏个数一定大于lcm/size;这个用反证法就可证明. 然后就是只需要枚举N%lcm + lcm这个体积的最有分配. */ #include<cstdio> #include<cstring> #incl…
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=4099 这个题目就是一个坑或. 题意:给你不超过40的一串数字,问你这串数字是Fibonacci多少的开头几位数字,如果不存在则输出-1. 题解:明明说好的不超过40,但是在建字典数的时候不加i<41就超内存了,杭电你是想咋地,害的我比较好多人的代码,一点一点试出来的. AC代码: #include <iostream> #include <cstdio> #include <…