POJ2965(找规律)题解 题目 The Pilots Brothers’ refrigerator Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 35766 Accepted: 13828 Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to
题目传送门 【题目大意】 有一个$4*4$的矩阵，现在要把矩阵内所有的“＋”变成“－”，已知如果选取一个位置$(i,j)$改变状态，那么第$i$行和第$j$列的所有位置的状态都会改变，求达到要求的最小步数和改变状态的顺序。 【思路分析】 分析可得，对于某个“＋”的位置$(i,j)$，要使它的状态改变而其他
#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int a[4][4];int x[17]= {0},X[17]= {0};int y[17]= {0},Y[17]= {0};int d=0,e=-1;void build(){ memset(a,0,sizeof(a)); for(int i=0; i<4; i++) { for
POJ 2965 The Pilots Brothers' refrigerator Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator. There are 16 handles on the refrigerator door. Every handle can
