链接:https://www.nowcoder.com/acm/contest/104/A来源:牛客网 题目描述 Let  be a regualr triangle, and D is a point in the triangle. Given the angle of . Then let AD, CD and BD form a new triangle, what is the size of the three angles?     #include<math.h> #inclu…

题目链接:http://acm.swust.edu.cn/problem/771/    Description 将军有一个用栅栏围成的矩形农场和一只奶牛,在农场的一个角落放有一只矩形的箱子,有一天将军要出门,他就把奶牛用一根绳子套牢,然后将绳子的另一端绑到了那个箱子不靠栅栏的角上,现在给定箱子的长和宽,绳子的长度,你的问题是计算奶牛能够到达的面积.   Input 有多组测试数据. 每一组数据为一行,三个整数,L(0<=L<=500),M,N(1<=M,N<=500),分别表示绳…

Faulhaber’s Triangle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 157    Accepted Submission(s): 78 Problem Description The sum of the m th powers of the first n integers S(n,m) = SUM ( j= 1…

2010浙江省数学竞赛,附加题. 设$D,E,F$分别为$\Delta ABC$的三边$BC,CA,AB$上的点,记$\alpha=\dfrac{BD}{BC},\beta=\dfrac{BD}{BC},\gamma=\dfrac{AF}{AB}$ 证明:$S_{\Delta DEF}\ge\alpha\beta\gamma S_{\Delta ABC}$ 证明:$S_{DEF}=S_{ABC}-S_{AFE}-S_{BDF}-S_{DCE}$$=S_{ABC}(1-\sum\limits_{c…

这一题题目有点坑,注意这句话: 这代表了其圆心就是矩形的中心! 然后就可以推公式: 可知: x = 200/(a+2atan(b/c)*r); r = sqrt(a*a + b*b); 所以有AC代码如下: #include <iostream> #include <cstdio> #include <cmath> using namespace std; int main() { ; double a, b; while(scanf("%lf : %lf&q…

A. Triangle 题目连接: http://codeforces.com/contest/6/problem/A Description Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to s…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1385 题意:下图中已知面积 a b c 求 d; 如果d的面积不确定,输出-1. 连接AX,记△AFX的面积为m,△AEX的面积为n. 因为△AFX与△BFX共高线,所以m/a = AF/FB; 又因为(n+c)与b共底边,高线所在的三角形相似,高线只比为AF/FB = (n+c)/b = m/a. 同理可得(m+a)/b = n/c = AE/CE 然后解二元一次方程组求得 m =…

King's Sanctuary Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/93 Description The king found his adherents were building four sanctuaries for him. He is interested about the positions of the sanctuaries and want…

题意:给定两个三角形,问你能不能拼成矩形. 析:很明显,要想是矩形,必须是四个角是直角,那么三角形必须是直角三角形,然后就是只能斜边相对,然后呢?就没了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <functional> #include <cstdlib> #inc…

(-￣▽￣)-* #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> using namespace std; struct node { double l,r; //找到以岛为圆心,以d为半径的圆与坐标x轴的左交点l.右交点r //雷达只有设在l~r之间,岛才在雷达覆盖范围内 }a[]; int cmp(node a,node b) { return a.l<…