HDU5512Pagodas 【gcd】

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本文介绍了一个基于数学博弈论的问题——两个和尚如何轮流重建古塔。通过分析剩余可重建古塔的数量，可以预测哪位和尚将在游戏中获胜。文章提供了一种算法实现方案，包括求最大公约数来确定可重建的古塔数量，并据此判断胜负。

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Pagodas

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1395    Accepted Submission(s): 974


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）

不难发现最终可以建 $n/gcd(a,b)-2$ 个塔，判断一下奇偶性即可

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include <string.h>
#include<math.h>

using namespace std;
#define ll long long
#define pii pair<int,int>

const int inf=1e9+7;
const int N = 2e3+5;

int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}

int main(int argc, char *argv[])
{
    //freopen("/home/lu/Documents/r.txt","r",stdin);
    int T;
    scanf("%d",&T);
    for(int t=1;t<=T;++t){
        int n,a,b;
        scanf("%d%d%d",&n,&a,&b);
        bool flag=(n/gcd(a,b))%2;
        printf("Case #%d: %s\n",t,flag?"Yuwgna":"Iaka");
    }
    return 0;
}