hdu5512 Pagodas (gcd求等差数列最小公差)

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

思路:

a+b或a-b这样多次之后，最后的序列其实是等差数列，
最开始的两个元素是等差数列中的元素，最小公差就是gcd(a,b)
n/gcd(a,b)就是序列中元素的个数，求出个数之后判断一下奇偶就能得出答案了。

code:

#include<bits/stdc++.h>
using namespace std;
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
int main(){
    int T;
    cin>>T;
    int cas=1;
    while(T--){
        int n,a,b;
        cin>>n>>a>>b;
        int d=n/gcd(a,b)-2;//这里减不减不影响奇偶性
        printf("Case #%d: ",cas++);
        if(d%2==0){
            cout<<"Iaka"<<endl;
        }else{
            cout<<"Yuwgna"<<endl;
        }
    }
    return 0;
}