F - Pagodas HDU - 5512

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 kk respectively, 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.

InputThe 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. OutputFor 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

題意：由題目中給出的兩個數相加或相減得到新的數，再將這些數繼續相加相減直到不再出現新的數。

題解：i1=j+k,i2=j-k,i3=2j,i4=3k...可以推出i=mj+nk，即i%gcd(j,k)==0,ans=n/gcd(j,k).

#include<iostream>
#include<algorithm>
#include<string.h>
#include<cmath>
#include<stdio.h>
using namespace std;
int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}
int main(){
    int cas;scanf("%d",&cas);
    for(int c=1;c<=cas;c++){
        int n,a,b;
        scanf("%d%d%d",&n,&a,&b);
        int d=gcd(a,b);
        int ans=n/d;
        printf("Case #%d: ",c);
        if(ans%2)
            printf("Yuwgna\n");
        else
            printf("Iaka\n");
    }
}