题意：给出9个时钟的状态和9种操作，每种操作可以改变不同的时钟的状态，每次只能改变90度，输出最少的有序的操作序列思路：枚举出每种操作的次数，最多为3次，保存在num数组中，判断每个时钟经过9种，每种num[j]次操作，是否为0.#include <stdio.h>#include <string.h>int flag;int state[9], num[9];int change[9][9]={{1, 1, 0, 1, 1, 0, 0, 0, 0}, {1, 1, 1, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 0, 1, 1, 0, 0, 0}, {1, 0, 0, 1 ,0, 0, 1, 0, 0}, {0, 1, 0, 1, 1, 1, 0, 1, 0}, {0, 0, 1, 0, 0, 1, 0, 0, 1}, {0, 0, 0, 1, 1, 0, 1, 1, 0}, {0, 0, 0, 0, 0, 0, 1, 1, 1}, {0, 0, 0, 0, 1, 1, 0, 1, 1}}; //9种操作，change[i][j]==1 表示第i种操作可以改变第j个时钟的状态。int judge(){for (int i=0; i<9; i++)//9个时钟{int tem=state[i];for (int j=0; j<9; j++) //9种操作{for (int k=0; k<num[j]; k++) //每种操作的次数{tem=(tem+change[j][i])%4;}}if (tem) return 0;}return 1;}void dfs(int x){if (x>9) return;if (judge()){flag=1;for (int i=0; i<9; i++){for (int j=0; j<num[i]; j++) printf("%d ", i+1);}printf("\n");return;}for (int i=0; i<=3; i++){num[x]=i;dfs(x+1);if (flag) return;}}int main(){//freopen("in.txt", "r", stdin);for (int i=0; i<9; i++) scanf("%d", &state[i]);memset(num, 0, sizeof(num));flag=0;dfs(0);return 0;}
