https://vjudge.net/contest/386568#problem/B

InputThe first line of the input gives the number of test cases,T(1≤T≤100)T(1≤T≤100).TTtest cases follow.

For each test case, the first line contains an integern(1≤n≤1000)n(1≤n≤1000), the number of weapons in the game.

Thennnlines follow, theithithof which contains two integersAi,Di(1≤Ai≤100,1≤Di≤10000)Ai,Di(1≤Ai≤100,1≤Di≤10000), representing the Damage and the Delay of each weapon.

The sum ofnnin all test cases doesn't exceed20002000.
OutputFor each test case, print a line with a real numberp(0≤p≤1)p(0≤p≤1), representing the optimal probability.

Your answers should have absolute or relative errors of at most10−610−6.
Sample Input

2
1
100 100
4
50 50
40 20
30 10
20 100

Sample Output

0.5
0.875

題意：

　　父親 和 兒子 對戰遊戲：

　　　　從n個武器中選擇一種，每種武器有傷害A，兩次使用的時間間隔D

　　　　父親的武器隨機選擇，兒子選一種武器，使得兒子獲勝概率最大。

　　　　輸出最大獲勝率

思路：

　　貪心，兒子選擇價效比最高的，使得在時間相同的情況下打出的傷害更快達到100；

　　父親選擇強道具的概率 = 強道具個數 / n；

　　則兒子獲勝概率為 (1/2)強道具個數 / n；

　　1）父親未選擇強武器，概率（n - 強道具個數）/強道具個數；

　　2）在此情況下兒子勝率為（n - 強道具個數）/ 強道具個數

　　最大概率為兩數和

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=2e5+10;
const int mod=1e9+7;


double P = 0;
double A = 0 , D = 0;
double Bestchoice = 0;
double T = 0;
double T_min = 1e9;
int main()
{
	int t = 0;
	int n = 0;
	cin >> t;
	while(t--)
	{
		Bestchoice = 0;
		T = 0;
		T_min = 1e9;
		cin >> n;
		for(int i = 0;i<n;i++)
		{
		 	cin >> A >> D;
		 	T = D * ((100+A-1)/A - 1);
		 	if(T == T_min)
			{
			 	Bestchoice += 1;
			} 
		 	if(T < T_min)
			{
		 	 	T_min = T;
		 		Bestchoice = 1;
			}
		}	
	    P = (n-Bestchoice)/n + Bestchoice/(n*2);
	    cout << setprecision(2) << P <<endl;
    }
	return 0;
}