多校第四场——1002 Blow up the Enemy

Problem Description
Zhang3 is playing a shooting game with Father. In the game there are two players trying to kill each other to win the game.

The game provides n weapons, each has two properties: Damage and Delay. The ith weapon has Damage Ai and Delay Di. When a player shoots with this weapon, his enemy’s HP is reduced by Ai, then he must wait for Di ms before he can shoot again.

The game processes as follows:

  1. Before the game starts, Zhang3 and Father choose a weapon respectively. Father always randomly chooses one of the n weapons with equal probabilities. Each player can only use the chosen weapon during the game.
  2. When the game starts, Zhang3 and Father have 100 HP each. They make their first shot at the same time.
  3. They keep shooting as quickly as possible. That means, a player shoots instantly whenever he can shoot, until the game ends.
  4. When a player’s HP is reduced to 0 or lower, he dies and the game ends. If the other player is still alive (i.e. has HP higher than 0), then the living player wins the game; otherwise (if the two players die at the same time), each player has 50% probability to win the game.

Zhang3 wants to win the game. Please help her to choose a weapon so that the probability to win is maximized. Print the optimal probability.

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

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

Then n lines follow, the ith of which contains two integers Ai,Di(1≤Ai≤100,1≤Di≤10000), representing the Damage and the Delay of each weapon.

The sum of n in all test cases doesn’t exceed 2000.

Output
For each test case, print a line with a real number p(0≤p≤1), representing the optimal probability.

Your answers should have absolute or relative errors of at most 10−6.

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

Sample Output
0.5
0.875

题意:两人对战,有n中武器可选,武器有伤害和冷却时间,一人随机选,另一人主动选,问主动选的人的获胜概率

思路:找出伤害最快超过或等于100的枪,统计器数量num,概率就是1-num/2.0/n

代码:

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        int m=1e6+5;
        int num=0;
        for(int i=0;i<n;++i)
        {
            double A,D;
            cin>>A>>D;
            int temp = (ceil(100/A)-1)*D;
            if(temp<m)
            {
                m=temp;
                num = 1;
            }
            else if(temp==m)
            {
                num++;
            }
        }
        cout<<1-num/2.0/n<<endl;
    }
}

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转载自blog.csdn.net/weixin_43330910/article/details/107773443