HDU 5512 (Pagodas) 2015 ACM/ICPC Asia Shenyang Division (subject number 1004)

Problem Description
N Pagodas were standing erect in Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 1 to N. H Owever, only two of them (labelled A and B, where 1≤a≠b≤n) withstood the "Test of".

Two monks, Yuwgna and Iaka, decide to make glories great again. They take turns to build pagodas and Yuwgna takes. 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, lab Elled J and K respectively, such that i=j+k or i=j−k. Each pagoda can is rebuilt twice.

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


Input
The contains an integer t (1≤t≤500) which is the number of test cases.
For the all test case, the the 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'll 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

Meaning

There are n temples that need mending for a long time, only two (marked as a,b) intact do not need to repair, there are two monks in turn to repair the N-2 temple, each monk can only repair a temple marked as I, and requirements I meet i=j+k or i=j-k, each temple can only be built once;

Among them J and K represent already built good temple, Yuwgna first, ask the last who cannot build who loses;

Like what:

Example 1, there are 3 temples, a=1,b=2, then Yuwgna can only repair a third, then Yuwgna win;

Example 2, there are 6 temples, numbered 1 2 3 4 5 6,a=2,b=3, then Yuwgna can only repair the first or fifth, if Yuwgna repaired the first, then Iaka can fix the fourth, and so on.

Ideas:

After reading, decisive game, and then find the law, If-else judge wrote five or six, made six or seven times, the results of the whole WA;

Finally, the bug has been adjusted to find that when the number of temples is 6 o'clock, if a=3,b=6, then Yuwgna no temple can be built;

Then added a judgment: processing b/a=2, each time can only repair a integer times, submit or WA;

Then find a bug: When the number of 12 o'clock, a=9,b=12, then 3, 6 can be taken;

Finally think of the gcd of A and B; gcd C of A and B, then the integer multiples of C can be taken; For example, there are 16 temples, a=8,b=12, and 4 and 16 can be taken;

The following is the AC code:

#include <stdio.h>
int f (int a,int b)
{                       //gcd
    int temp,r;
    if (a<b)
    {
        temp=a;
        a=b;
        b=temp;
    }
    r=a%b;
  while (R)
  {
      a=b;
      B=r;
      r=a%b;
  }
  return b;
}
int main ()
{
    int t;
    int n,i,a,b;
    int j,sum;
    int k;
    scanf ("%d", &t);
    for (i=1;i<=t;i++)
    {
        k=0;
        scanf ("%d%d%d", &n,&a,&b);
        Sum=f (a,b);
        for (j=sum;j<=n;j+=sum)     //j each time the GCD k++ of A and B are added
            ;
        k-=2;
    if (k%2==0)
        printf ("Case #%d:iaka\n", i);
    else
        printf ("Case #%d:yuwgna\n", i);
    }
    return 0;
}