2021-02-01

最新推荐文章于 2022-05-06 22:25:22 发布

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D - Pagodas
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" orIaka"). 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

解题：找规律，所能枚举的数均为gcd（a，b）的倍数，以n/gcd（a，b）奇偶性判定胜负

详细证明：整体思路（所建造的山为gcd的倍数，并且所有的倍数可以被枚举出）

无论如何搭配都是qa+pb的形式为gcd（a，b）的倍数
利用 gcd(a,b)=gcd(b,a-b) 可以一步步推出最大公约数x（推导过程即建造过程，可以看下面网站实例）
可参考https://baike.baidu.com/item/%E6%9B%B4%E7%9B%B8%E5%87%8F%E6%8D%9F%E6%9C%AF/449183?fromtitle=%E6%9B%B4%E7%9B%B8%E5%87%8F%E6%8D%9F%E6%B3%95&fromid=10277459&fr=aladdin
不妨设b！=x 则b=sx，利用x和sx可以推出（s-1）x，类似的可轻易推出x，2x，3x····

#include<iostream>
using namespace std;
int gcd(int a,int b)//上文中更相减损法仅为了过程推导，在此使用了辗转相除法
{
    while(b) b^=a^=b^=a%=b;
    return a;
}
int main()
{
    int t;
    cin>>t;
    int con=0;
    while(t--)
    {
        con++;
        int n,a,b;
        int ans=0;
        cin>>n>>a>>b;
        int r=gcd(a,b);
        ans=n/r-2;
        cout<<"Case #"<<con<<": ";
        if(ans%2) cout<<"Yuwgna"<<endl;
        else cout<<"Iaka"<<endl;
    }
}