Doors Breaking and Repairing（补题）解题报告

题目描述：

You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consists of two phases. During the first phase you make your move and during the second phase Slavik makes his move.

There are nn doors, the ii-th door initially has durability equal to aiai.

During your move you can try to break one of the doors. If you choose door ii and its current durability is bibi then you reduce its durability to max(0,bi−x)max(0,bi−x) (the value xx is given).

During Slavik’s move he tries to repair one of the doors. If he chooses door ii and its current durability is bibi then he increases its durability to bi+ybi+y (the value yy is given). Slavik cannot repair doors with current durability equal to 00.

The game lasts 1010010100 turns. If some player cannot make his move then he has to skip it.

Your goal is to maximize the number of doors with durability equal to 00 at the end of the game. You can assume that Slavik wants to minimize the number of such doors. What is the number of such doors in the end if you both play optimally?

Input：

The first line of the input contains three integers nn, xx and yy (1≤n≤1001≤n≤100, 1≤x,y≤1051≤x,y≤105) — the number of doors, value xx and value yy, respectively.

The second line of the input contains nn integers a1 ,a2,…,ana1,a2,…,an (1≤ai≤1051≤ai≤105), where aiai is the initial durability of the ii-th door.

Output：

Print one integer — the number of doors with durability equal to 00 at the end of the game, if you and Slavik both play optimally.

Sample Input：

6 3 2
2 3 1 3 4 2

Sample Output：

6

Sample Input：

5 3 3
1 2 4 2 3

Sample Output：

2

Sample Input：

5 5 6
1 2 6 10 3

Sample Output：

2

题目大意：

一个游戏有2个人，A,B，按照回合制，面前有N扇们，每个回合分为2个阶段，A先走，他遇到门可以破坏他的耐久度，最大为x（题目会给），B后走，他遇到门可以修复到当前耐久度加y。问当游戏结束，有多少耐久度为0的门。

思路分析：

 1. 当x>y的时候，无论B如何修复，都会被A给破坏，所以游戏
 结束的时候，全部门都会破坏
 2. 当x<=y的时候，假设有5扇门，有3扇可以被A破坏，那么A破
 坏了其中一扇，到了B回合，他是不能破坏已经被A破坏成0的门，
 那么他可以去修复，3扇门其中一扇把他提高耐久度，让A破坏不
 了，那么这里找到规律，当x<=y的时候，所能破坏的门/2，当奇
 数的时候还要多加1，因为回合是A先，所以A肯定可以多破坏一个

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	int n,x,y;
	while(cin >> n >> x >> y)
	{
		int num=0,a;
		for(int i=0;i<n;i++)
		{
			cin >> a; 
			if(x>=a)
			{
				num++;
			}
		}
		if(x>y)
		{
			cout << n << endl;
		}
		else if(x<=y)
		{
			if(num%2==0)
			{
				cout << num/2 << endl; 
			}
			else
			{
				cout << num/2+1 << endl;
			}
		}
	}
 } 



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