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 << endlconst 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;}