脑洞博弈 E. Competitive Seagulls 2017 ACM Arabella Collegiate Programming Contest

题目链接：http://codeforces.com/gym/101350/problem/E

题目大意:给你一个长度为n的方格，方格上面都被染色成了白色。每次染色都是选择白色的，假设目前选择的这块白色（白色联通块）的长度为L，每次都只能选择<=(L+1)/2的素数染色。问谁赢？

思路：果然博弈不会啊，感谢这位神牛：链接

当n=2||n=3先手必败

其余的n必胜，n=1就不解释了。当n>=4的时候，如果是奇数，那么每次都取出中间两个，如果是偶数，每次都取出中间三个，然后对称取即可。

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")int main(){int t; cin >> t;while (t--){int n; scanf("%d", &n);if (n == 2 || n == 3) puts("second");else puts("first");}return 0;
}

View Code

转载于:https://www.cnblogs.com/heimao5027/p/6827538.html

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