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2015ACM-ICPC亚洲区域赛沈阳站 D：Pagodas
source link: https://arminli.com/acm-shenyang-d/
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Armin's Blog
2015ACM-ICPC亚洲区域赛沈阳站 D：Pagodas
March 09, 2016
题意：N 个点，刚开始给出两个点 a，b(a != b) ，有两个人玩一个游戏，游戏规则如下：每次只能选择 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 。问最后谁一个点也选不了了。
能被选的点其实只有 n / GCD(a,b)，因为初始的 a 和 b 决定了塔的间距，最后只要判奇偶即可。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int gcd(int a, int b){
return b == 0 ? a : gcd(b, a % b);
}
int main(){
int cas;
scanf("%d", &cas);
for (int t = 1; t <= cas; t++) {
int n, a, b;
scanf("%d%d%d", &n, &a, &b);
int d = n / gcd(a, b);
printf("Case #%d: %s\n", t, d&1 ? "Yuwgna" : "Iaka");
}
return 0;
}
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