2015ACM/ICPC亚洲区沈阳站-——Pagodas（简单博弈）

nn pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 11 to nn. However, only two of them (labelled aa and bb, where 1≤a≠b≤n1≤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)i (i∉{a,b} and 1≤i≤n) if there exist two pagodas standing erect, labelled jj and kkrespectively, such that i=j+ki=j+k or i=j−ki=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)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)n (2≤n≤20000) and two different integers aa and bb.

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

 

解题思路：

      试过两组数据后，可以很清楚看到：最左边的一个可修建位置一定是gcd(a,b) !!! 并且所有能修建的点都一定是这个gcd的整数倍。那么....总共能修建ans =  n/gcd(a,b)次，  根据奇偶性，奇数先手的Yuwgna获胜。

#include
using namespace std;
int main(void)
{
     int T;cin>>T;
     int cas = 0; 
     while(T--)
     {
        int n,a,b; 
        cin>>n>>a>>b;
        int gc = __gcd(a,b); 
        int tot = n/gc;    
     	printf(tot&1 ?"Case #%d: Yuwgna\n":"Case #%d: Iaka\n",++cas);   
     }
}