Problem ATriangle Fun Input: Standard Input Output: Standard Output In the picture below you can see a triangle ABC. Point D, E and F divides the sides BC, CA and AB into ratio 1:2 respectively. That is CD=2BD, AE=2CE and BF=2AF. A, D; B, E and C, F…
描述 Background Hello Earthling. We're from the planet Regetni and need your help to make lots of money. Maybe we'll even give you some of it. You see, the problem is that in our world, everything is about integers. It's even enforced by law. No other…
Herding Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1125 Accepted Submission(s): 325 Problem Description Little John is herding his father's cattles. As a lazy boy, he cannot tolerate cha…
Herding Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 702 Accepted Submission(s): 174 Problem Description Little John is herding his father's cattles. As a lazy boy, he cannot tolerate cha…
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32071 Accepted Submission(s): 16490 Problem Description " 改革春风吹满地, 不会AC没关系; 实在不行回老家, 还有一亩三分地. 谢谢!(乐队奏乐)" 话说部分学生心态极好,每天就知道游戏,这次考试如此简单的题目,也是云里雾里,…
代码如下: func GetTriangleAreaByVector(x vector.Vector3,y vector.Vector3,z vector.Vector3) float64 { //根据三角形三个点坐标求面积 //先算三角形三个边的长度 a := vector.GetDistance(x,y) b := vector.GetDistance(x,z) c := vector.GetDistance(y,z) s := (a + b + c) / 2 area := math.Sq…
AC代码: #include<cstdio> #include<cmath> #include<algorithm> #include<iostream> #include<cstring> using namespace std; typedef long long ll; ; int sgn(double x) { ; ? - : ; } int main() { double a, b ,c; while(~scanf("%lf…
Problem Description This is a simple problem. Given two triangles A and B, you should determine they are intersect, contain or disjoint. (Public edge or point are treated as intersect.) Input First line contains an integer T (1 ≤ T ≤ 10), represents…
""" 已知三角形的边长求他的面积和周长 Author:罗万财 Date:2017-3-3 """ import math a=float(input('a=')) b=float(input('b=')) c=float(input('c=')) if a+b>c and a+c>b and b+c>a: d=a+b+c e=(a+b+c)/2 f=math.sqrt(e*(e-a)*(e-b)*(e-c)) print('三…
