【Hrbust】 Alice and Bob (博弈)

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本文介绍了一个结合数论与博弈论的游戏“A和B”，详细解析了游戏规则及获胜策略。通过计算特定整数N下可能的状态数，利用记忆数组避免重复计算，最终判断先手玩家Alice是否能赢。

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A.Alice and Bob

Time Limit: 1000 MS

Memory Limit: 32768 K

Total Submit: 150 (37 users)

Total Accepted: 31 (29 users)

Special Judge: No

Description

Alice and Bob are playing a game called A and B, and the loser will do homework for the winner. The name of the game doesn't come from the initial character
 of their names. The game is called A and B because they need to write down integer A and integer B in turns and A2 - B2 must be equal to N. N is also an integer given by Jim Zheng. He is the judge of the game. Every time Alice or Bob
 can write at most 3 sets of A and B, and at least 1 set if there still exist A and B (A2 - B2 = N). The person who can’t write A and B in his/her turn will be the loser. In the game, Alice always writes
 A and B first.
For example, Jim Zheng shows an integer: 5 to Alice and Bob. First, Alice writes 3 sets of A and B: (3, 2), (3, -2), (-3, 2). Then Bob writes 1 set of A
 and B: (-3, -2). Now, it’s Alice’s turn again. But she can’t write A and B any more, so she loses.
Now, we can assume Alice and Bob both use the best strategy. Giving you an integer N. Can you tell Alice whether she can win by using the best strategy?
 Please output “yes” if Alice can win, or “no” if she can’t.

Input

First line contains an integer T (T >= 1000000), which is the number of cases.
In the next T lines, every line contains an integer N (1 <= N <= 1000000).

Output

Just output one line for each case, output “yes” if Alice can win, or “no” if she can’t.

Sample Input

Sample Output

yes
no
no

Author

Zheng Jintao

一个简单的数论和博弈混合的题目。

在算出每个答案后，要用记忆数组保存一下，防止相同数据多次计算。

代码：

#include <cstdio>
#include <cmath>
#include <climits>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#define LL long long
#define db double
#define pi acos(-1.0)
#define pr printf
#define sc scanf
#define mod
#define N 1000111
using namespace std;

int c[N];

int main()
{
    int T,i,j,n,a,b,sum;

    for(i=1; i<N; ++i)
        c[i]=-1;

    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);

        sum=0;

        if(c[n]!=-1)
            sum = c[n];
        else
        {
            for(i=1; i*i<=n; ++i)
            {
                if(n%i==0&&((n/i-i)%2==0))
                {
                    if(i==n/i)
                        sum +=2;
                    else
                        sum+=4;
                }
            }
            c[n] = sum;
        }
        puts((sum%4)?"yes":"no");
    }
    return 0;
}