# 山东省第六届ACM省赛题——BIGZHUGOD and His Friends II（塞瓦定理

2016-12-02 12:52:44来源:网络收集作者:管理员人点击

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.

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).

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”.

2
1 1 1
3 4 6

YES 0.5000
YES 2.0000

#include #include #include #include #include #include #include #include #include #include #include #define MAXN 100010 #define mod 9973 #define INF 0x3f3f3f3f using namespace std; int main() { ios::sync_with_stdio(false); int t,t1,t2,t3; cin>>t; while(t--) { cin>>t1>>t2>>t3; double low=0,high=min(t1,min(t2,t3)),mid; double f0; do { mid=(low+high)*0.5; f0=(t1-mid)*(t2-mid)*(t3-mid)-mid*mid*mid; if(f0>0) low=mid; else if(f0<0) high=mid; }while(fabs(f0)>1e-9); cout<<"YES "<