【Gym - 101350 E - Competitive Seagulls】 & 博弈

最新推荐文章于 2019-07-26 16:04:54 发布

楚江枫

最新推荐文章于 2019-07-26 16:04:54 发布

阅读量863

点赞数

分类专栏：博弈论

本文链接：https://blog.youkuaiyun.com/WYK1823376647/article/details/77460578

版权

codeforces 同时被 2 个专栏收录

172 篇文章

订阅专栏

博弈论

14 篇文章

订阅专栏

本文介绍了一种由两只海鸥玩的特殊游戏策略。游戏中，玩家需在一排初始全为白色的方格中，选择一定长度的连续白色方格并将其涂黑。玩家的目标是在对手无法行动时赢得比赛。文章通过分析得出结论，在大多数情况下，先手玩家将获胜。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

There are two seagulls playing a very peculiar game. First they line up N unit squares in a line, all originally colored white.

Let L be the length of the longest continuous sub-segment of white unit squares. Let P be any prime that satisfies the condition that . The player can choose any P if it satisfies the conditions but if there is no value of P that does, then P is set to 1.

The current player then proceeds to choose any continuous sub-segment of white unit squares of length P and paint them black. The player who can’t make a move loses.

If both players play optimally and the first player starts, which player is going to win the game?

Input
The first line of input is T – the number of test cases.

Each test case contains a line with a single integer N (1 ≤ N ≤ 107).

Output
For each test case, output on a single line “first” (without quotes) if the first player would win the game, or “second” if the second would win.

Example
Input
2
2
5
Output
second
first

题意：给出一行 1 * n 的格子，一开始全是白色的，你可以选择不大于 p <=[L / 2] (向上取整，L为当前连着的最大数量的白色格子)的质数（没有不大于 p 的质数时，p取 1）去填任意 p 个格子，问最后先手和后手，谁最后一个把格子填满

思路：模拟几个数会发现，只用在 n 等于 2 或 3 时后手会赢，其余都是先手赢

AC代码：

#include<cstdio>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e5 + 10;
typedef long long LL;
int main()
{
    int T; scanf("%d",&T);
    while(T--){
        int n,a;
        scanf("%d",&a);
        if(a == 2 || a == 3) puts("second");
        else puts("first");
    }
    return 0;
}