HDU 5512 Pagodas （gcd）

题意:

n<=2∗104座塔,一开始只有a,b塔,a!=b<=n,对于已建成的塔j，k(包括a，b两塔),每次只能重建编号为j+k或j−k的塔,Yuwgna和Iaka轮流建塔，Yuwgna先手，最后不能建的人输，问最终获胜的人是哪个

分析:

更相减损,可知就是gcd,所以能建立的塔总数是n/gcd(a,b),去掉刚开始的2座,判断下奇偶就好了

代码:

//
//  Created by TaoSama on 2015-10-31
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, a, b;

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
while(t--) {
scanf("%d%d%d", &n, &a, &b);
int cnt = n / __gcd(a, b) - 2;
printf("Case #%d: %s\n", ++kase, cnt & 1 ? "Yuwgna" : "Iaka");
}
return 0;
}