Pagodas

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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）

题意：1到n个塔，Yuwgna和Iaka两人从a，b两点开始修建塔，每次可以修建塔的位置i=j+k或i=j-k；j和k为已修建好的塔的位置，Yuwgna开始修，最后谁不能修了谁就输了，问最后谁赢。

分析：最简单的签到题；首先判断a和b中有没有1，如果有的话，那么每个位置都能修建塔；如果没有，在判断a和b是不是互质，如果互质，那么肯定会出现1的情况，也就是1这位置总有一回合会修建塔，所以也是每个位置都能修建塔，然后就是不互质的情况了，不互质的话，可以修建塔的位置的个数就只有n/gcd(a, b)；然后最后判断总共可以修建的塔的个数是奇数还是偶数就ok了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))

int main()
{
    int T,n,a,b;
    cin>>T;
    for(int cas=1; cas<=T; cas++)
    {
        cin>>n>>a>>b;
        if(b<a) swap(a, b);
        cout<<"Case #"<<cas<<": ";
        int t=__gcd(a, b);//a和b的最大公约数
        int cnt;
        if(a==1)//有1
            cnt=n;
        else
        {
            if(t!=1)//不互质
                cnt=n/t;
            else
                cnt=n;
        }
        if(cnt&1)
            cout<<"Yuwgna"<<endl;
        else
            cout<<"Iaka"<<endl;
    }
    return 0;
}

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