HDU 5512 Meeting 博弈论

Meeting

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

Description

n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 1 to n. However, only two of them (labelled a and b, where 1≤a≠b≤n) withstood the test of time.

Two monks, Yuwgna and Iaka, decide to make glories great again. They take turns to build pagodas and Yuwgna takes first. For each turn, one can rebuild a new pagodas labelled i (i∉{a,b} and 1≤i≤n) if there exist two pagodas standing erect, labelled j and k respectively, such that i=j+k or i=j−k. Each pagoda can not be rebuilt twice.

This is a game for them. The monk who can not rebuild a new pagoda will lose the game.

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

16
2 1 2
3 1 3
67 1 2
100 1 2
8 6 8
9 6 8
10 6 8
11 6 8
12 6 8
13 6 8
14 6 8
15 6 8
16 6 8
1314 6 8
1994 1 13
1994 7 12

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