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Meeting

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5512

Description

Input

The first line contains an integer t (1≤t≤500) which is the number of test cases.
For each test case, the first line provides the positive integer n (2≤n≤20000) and two different integers a and b.

Output

For each test case, output the winner (``Yuwgna" or ``Iaka"). Both of them will make the best possible decision each time.

Sample Input

Sample Output

Case #1: Iaka
Case #2: Yuwgna
Case #3: Yuwgna
Case #4: Iaka
Case #5: Iaka
Case #6: Iaka
Case #7: Yuwgna
Case #8: Yuwgna
Case #9: Iaka
Case #10: Iaka
Case #11: Yuwgna
Case #12: Yuwgna
Case #13: Iaka
Case #14: Yuwgna
Case #15: Iaka
Case #16: Iaka

HINT

题意

有n个地点，然后在a,b位置已经修建了佛塔了，每个地点，只能修建一次佛塔

然后有两个人依次修建佛塔

如果要在x位置修建佛塔，那么需要满足存在i,j位置已经修建了佛塔，且x = i + j 或者 x = i - j

谁不能修建，就谁输了

让你输出最后谁能够胜利

题解：

这是一个博弈论，看到a-b的时候，就请想到gcd……

大胆猜一猜结论…… n/gcd(a,b)的奇数和偶数

代码

#include<iostream>
#include<stdio.h>
using namespace std;

int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    int t;scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        int a,b,n;
        cin>>n>>a>>b;
        int flag = n/gcd(a,b);
        if(flag%2==0)
            printf("Case #%d: Iaka\n",cas);
        else
            printf("Case #%d: Yuwgna\n",cas);
    }
}

HDU 5512 Meeting 博弈论

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原文地址：http://www.cnblogs.com/qscqesze/p/4929923.html