Vasya and Chess - CodeForces 493 D 博弈

最新推荐文章于 2020-07-29 09:55:01 发布

提比-我有特殊的AC技巧

最新推荐文章于 2020-07-29 09:55:01 发布

阅读量756

点赞数

CC 4.0 BY-SA版权

分类专栏： CodeForces 博弈论

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/41720563

CodeForces 同时被 2 个专栏收录

82 篇文章

订阅专栏

博弈论

29 篇文章

订阅专栏

在n*n的棋盘上，白皇后位于(1,1)，黑皇后位于(1,n)，其余格子布满中立棋子。双方轮流行动，必须吃掉一枚棋子，无法行动者判负。本文分析了不同棋盘大小下的最优策略。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Vasya decided to learn to play chess. Classic chess doesn't seem interesting to him, so he plays his own sort of chess.

The queen is the piece that captures all squares on its vertical, horizontal and diagonal lines. If the cell is located on the same vertical, horizontal or diagonal line with queen, and the cell contains a piece of the enemy color, the queen is able to move to this square. After that the enemy's piece is removed from the board. The queen cannot move to a cell containing an enemy piece if there is some other piece between it and the queen.

There is an n × n chessboard. We'll denote a cell on the intersection of the r-th row and c-th column as (r, c). The square (1, 1)contains the white queen and the square (1, n) contains the black queen. All other squares contain green pawns that don't belong to anyone.

The players move in turns. The player that moves first plays for the white queen, his opponent plays for the black queen.

On each move the player has to capture some piece with his queen (that is, move to a square that contains either a green pawn or the enemy queen). The player loses if either he cannot capture any piece during his move or the opponent took his queen during the previous move.

Help Vasya determine who wins if both players play with an optimal strategy on the board n × n.

Input

The input contains a single number n (2 ≤ n ≤ 109) — the size of the board.

Output

On the first line print the answer to problem — string "white" or string "black", depending on who wins if the both players play optimally.

If the answer is "white", then you should also print two integers r and c representing the cell (r, c), where the first player should make his first move to win. If there are multiple such cells, print the one with the minimum r. If there are still multiple squares, print the one with the minimum c.

Sample test(s)

input
2

output
white
1 2

input
3

output
black

Note

In the first sample test the white queen can capture the black queen at the first move, so the white player wins.

In the second test from the statement if the white queen captures the green pawn located on the central vertical line, then it will be captured by the black queen during the next move. So the only move for the white player is to capture the green pawn located at (2, 1).

Similarly, the black queen doesn't have any other options but to capture the green pawn located at (2, 3), otherwise if it goes to the middle vertical line, it will be captured by the white queen.

During the next move the same thing happens — neither the white, nor the black queen has other options rather than to capture green pawns situated above them. Thus, the white queen ends up on square (3, 1), and the black queen ends up on square (3, 3).

In this situation the white queen has to capture any of the green pawns located on the middle vertical line, after that it will be captured by the black queen. Thus, the player who plays for the black queen wins.

题意：在n*n的棋盘上，(1,1)点为白皇后，(1,n)点为黑皇后，其他位置布满中立的棋子，由白色先手，当一个人没有棋子可以吃的时候，或是自己的棋子被对方吃掉之后，那么它就输了，问最后是谁赢。

思路：首先看当n是奇数的时候，我们可以把最中间的一列看做对称线，那么白棋无论怎么走，黑棋都走一样的，那么最后会是白棋先走到这个对称线上，所以黑棋胜。当n为偶数的时候，只要白棋先走（1,2）这个点，那么又变成了上面的情况，先手的胜负判定同上。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
int main()
{
    int T,t,n,m,i,j,k;
    scanf("%d",&n);
    if(n&1)
    {
        printf("black\n");
    }
    else
       printf("white\n1 2\n");

}