D - Pagodas （找规律）

最新推荐文章于 2019-11-03 20:51:03 发布

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本文介绍了一个基于数学原理的游戏策略问题，玩家通过轮流向集合中添加符合条件的新元素来扩展集合，目标是在有限范围内最大化集合元素数量。游戏胜负由最终剩余位置的奇偶性决定。

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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

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

题意：给你三个数：n,a,b，一开始集合里面有两个数：a和b，然后两个人轮流往这个集合里面增加数字，增加的这个数字的原则是：这个集合里面任选两个数的和或差（a + b或a - b或b -a的中的任意一个没被选中的符合[1,n]的点），集合里面的数字不能重复，同时这个数字不能大于 n ，求最后哪个人选不了满足条件的数了。

解题思路：能被选的点其实只有n / GCD(a,b)，为什么呢，思考一下：a + b或a - b或b -a这样的数次操作之后，无论怎样，最后得到的这个集合里面的数列，其实是一个等差数列，因为初始的a和b决定了塔的间距，最后只要判奇偶即可。

贴代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <set>
#include <sstream>
#include <stack>
using namespace std;
typedef long long ll;
const int maxn = 1005;
int gcd(int a,int b){
    return b == 0 ? a : gcd(b,a%b);
}

int main()
{
    int t,n,a,b;
    int kase = 1;
    cin>>t;
    while(t--){
        cin>>n>>a>>b;
        int m = gcd(a,b);
        cout<<"Case #"<<kase++<<": ";
        if(n / m % 2 == 1){
            cout<<"Yuwgna"<<endl;
        }else{
            cout<<"Iaka"<<endl;
        }
    }
    return 0;
}