HDU 5512 Pagodas（思维 数学）

Pagodas

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2555    Accepted Submission(s): 1725

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  aand  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

 

如题求最后动不了的就输，那么求可以移动的步数即可以10 6 8为例 只能修复的房子为 4 6 8 10 2 即可知道6 8的步数等于10/gcd（6,8）即可

#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main(int argc, char *argv[])
{
    int t;
    int case1=1;
    scanf("%d",&t);
    while(t--)
    {
        int all,s,e;
        scanf("%d %d %d",&all,&s,&e);
        int gcd=__gcd(s,e);
        if((all/gcd)%2==0)
        {
            printf("Case #%d: Iaka\n",case1++);
        }
        else
        {
            printf("Case #%d: Yuwgna\n",case1++);
        }
    }
    return 0;
}