BIGZHUGOD and His Friends II

  1000 ms         65536 KiB Submit Status My Status  Origin

Description

BIGZHUGOD and his three friends are playing a game in a triangle ground.

The number of BIGZHUGOD is 0, and his three friends are numbered from 1 to 3. Before the game begins, three friends stand on three vertices of triangle in numerical order (1 on A, 2 on B, 3 on C), BIGZHUGOD stands inside of triangle.

Then the game begins, three friends run to the next vertex in uniform speed and in straight direction (1 to B, 2 to C, 3 to A and there speeds may different). And BIGZHUGOD can stand in any position inside the triangle.

When any of his friends arrives at next vertex, the game ends.

BIGZHUGOD and his friends have made an agreement: we assume that the beginning is time 0, if during the game, you can find a moment that BIGZHUGOD can block the sight line of 1 to C, 2 to A, 3 to B. Then each friend has to treat BIGZHUGOD with a big meal.

Now BIGZHUGOD knows the lengths of time that his three friends need run to next vertices t1, t2 and t3. He wants to know whether he has a chance to gain three big meals, of course he wants to know in which exciting moment t, he can block three friends\' sight line.

Input

The first line contains an integer T, indicating the number of test cases (T ≤ 1000).

For each case there are three integer t1, t2, t3 (1 ≤ t1, t2, t3 ≤ 100).

Output

If BIGZHUGOD has a chance to gain big meal from his friends, output "YES" and the exciting moment t rounding to 4 digits after decimal point. Otherwise, output "NO".

Sample

Input

Copy2 1 1 1 3 4 6

Output

CopyYES 0.5000 YES 2.0000

Source

“浪潮杯”山东省第六届ACM大学生程序设计竞赛 Submit Status My Status  Origin

鬼知道的

塞瓦定理

 

没见过    涨姿势了  https://baike.baidu.com/item/塞瓦定理/2675177?fr=aladdin

就是三个人站在了一个三角形的三个端点，问你有没有可能使三个人的目光交与一点，有这个定理可以看出来肯定可以，那就二分呗，主要是精度问题。

#include<bits/stdc++.h> using namespace std; int main() { int t; scanf("%d",&t); while(t--) { double t1,t2,t3; scanf("%lf%lf%lf",&t1,&t2,&t3); double high=min(t1,min(t2,t3)); double low=0; double mid; double f; do{ mid=(high+low)*0.5; // printf("mid:%lf\n",mid); f=(t1-mid)*(t2-mid)*(t3-mid)-mid*mid*mid; // printf("~%lf\n",f); if(f>0.0) low=mid; else if(f<0) high=mid; }while(abs(f)>1e-9); printf("YES %.4lf\n",mid); } return 0; }