首页 > ACM题库 > HDU-杭电 > HDU 4371-Alice and Bob-博弈论-[解题报告]HOJ
2015
05-24

HDU 4371-Alice and Bob-博弈论-[解题报告]HOJ

Alice and Bob

问题描述 :

Alice and Bob are interested in playing games. One day, they invent a game which has rules below:
1. Firstly, Alice and Bob choose some random positive integers, and here we describe them as n, d1, d2,…, dm.
2. Then they begin to write numbers alternately. At the first step, Alice has to write a “0”, here we let s1 = 0 ; Then, at the second step, Bob has to write a number s2 which satisfies the condition that s2 = s1 + dk and s2 <= n, 1<= k <= m ; From the third step, the person who has to write a number in this step must write a number si which satisfies the condition that si = si-1 + dk or si = si-1 – dk , and si-2 < si <= n, 1 <= k <= m, i >= 3 at the same time.
3. The person who can’t write a legal number at his own step firstly lose the game.
Here’s the question, please tell me who will win the game if both Alice and Bob play the game optimally.

输入:

At the beginning, an integer T indicating the total number of test cases.
Every test case begins with two integers n and m, which are described before. And on the next line are m positive integers d1, d2,…, dm.
T <= 100;
1 <= n <= 106;
1 <= m <= 100;
1 <= dk <= 106, 1 <= k <= m.

输出:

At the beginning, an integer T indicating the total number of test cases.
Every test case begins with two integers n and m, which are described before. And on the next line are m positive integers d1, d2,…, dm.
T <= 100;
1 <= n <= 106;
1 <= m <= 100;
1 <= dk <= 106, 1 <= k <= m.

样例输入:

2
7 1
3
7 2
2 6

样例输出:

Case #1: Alice
Case #2: Bob

/*
分析:
    简单的博弈,挺有意思的:
    最优策略是一直加最小的数;
    否则的话、对手就可以减去最小的数使得整体仍然是上升的,
酱紫的话、他就使对手可以使用减号、使对手妥妥的不败。

                                                        2013-07-17
*/

#include"iostream"
#include"cstdio"
#include"cstring"
using namespace std;

int main()
{
	int T,Case;
	int i;
	int n,m;
	int temp,min;
	cin>>T;
	for(Case=1;Case<=T;Case++)
	{
		cin>>n>>m;
		min=1<<30;
		for(i=0;i<m;i++)
		{
			scanf("%d",&temp);
			if(temp<min)	min=temp;
		}
		temp=n%(min<<1);
		printf("Case #%d: %s\n",Case,temp<min ? "Alice" : "Bob");
	}
	return 0;
}

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参考:http://blog.csdn.net/ice_crazy/article/details/9350073