HDU-5512 Pagodas（找最大公约数）

Pagodas Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3075    Accepted Submission(s): 2071   Problem 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   Source 2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）   Recommend wange2014

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int gcd(int a,int b)
{
    if(b==0)
    return a;
    else return gcd(b,a%b);
}
int main()
{
    int t;
    int cnt=0;
    scanf("%d",&t);
    while(t--)
    {
        int n;int a,b;
        scanf("%d%d%d",&n,&a,&b);
        printf("Case #%d: ",++cnt);
        swap(a,b);
        int xxx=gcd(a,b);
            if((n/xxx)%2==1)  printf("Yuwgna\n");
            else             printf("Iaka\n");
    }
    return 0;
}