Alice and Bob are playing a kind of special game on an N*M board (N rows, M columns). At the beginning, there are N*M coins in this board with one in each grid and every coin may be upward or downward freely. Then they take turns
to choose a rectangle (x 1, y 1)-(n, m) (1 ≤ x 1≤n, 1≤y 1≤m) and flips all the coins (upward to downward, downward to upward) in it (i.e. flip all positions (x, y) where x1≤x≤n, y 1≤y≤m)). The
only restriction is that the top-left corner (i.e. (x 1, y1)) must be changing from upward to downward. The game ends when all coins are downward, and the one who cannot play in his (her) turns loses the game. Here's the problem: Who
will win the game if both use the best strategy? You can assume that Alice always goes first.
Then T cases follow, each case starts with two integers N and M indicate the size of the board. Then goes N line, each line with M integers shows the state of each coin, 1<=N,M<=100. 0 means that this coin is downward in the initial, 1 means that this coin is upward in the initial.
2 2 2 1 1 1 1 3 3 0 0 0 0 0 0 0 0 0
Alice Bob
找一找就有规律了。
#include <cstdio> int main() { int N; scanf("%d",&N); while(N--) { int n , m , a; scanf("%d %d",&n , &m); for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) scanf("%d",&a); if(a) puts("Alice"); else puts("Bob"); } return 0; }

本文介绍了一种基于特殊规则的棋盘游戏,玩家通过翻转棋盘上的硬币来博弈,目标是使所有硬币朝下。文章提供了一个简单的算法实现,用于判断在最优策略下哪位玩家能够赢得游戏。
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