時間限制: 1 Sec  記憶體限制: 128 MB 提交: 55  解決: 36 [提交] [狀態] [討論版] [命題人:admin]

題目描述

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 6= 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 (1 ≤ i ≤ n) if there exist two pagodas standing erect, labelled j and k respectively, such that i = j + k or i = j − k. This is a game for them. The monk who can not rebuild a new pagoda will lose the game.

輸入

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.

輸出

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

樣例輸入

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

樣例輸出

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

#include<cstdio>
using namespace std;

int gcd(int a,int b)
{
    int r;
    while(b>0)
    {
        r=a%b;
        a=b;
        b=r;
    }
    return a;
}

int main()
{
    int n,a,b,t,c=0;
    scanf("%d",&t);
    while(t--&&scanf("%d%d%d",&n,&a,&b))
    {
        c++;
        if((n/gcd(a,b))%2==1)
             printf("Case #%d: Yuwgna\n",c);
        else  printf("Case #%d: Iaka\n",c);
    }
    return 0;
}