#include <iostream>#include <cmath>#include <cctype>#include <string>#include <map>#include <set>#include <vector>#include <algorithm>#include <list>#include <stack>#include <cstring>using namespace std;void operate(unsigned operations[10][10], unsigned clocks2[10], int op_num, int op_count){ for (int i = 1; i <= 9; i++) { clocks2[i] += operations[op_num][i]*op_count; clocks2[i] %= 4; }}int main(){ //保存原始的时钟状态 unsigned clocks[10] = {0}; //保存最后最少的移动次数，最多也就是27次 int min_res = 28, tmp = 0; //保存最小次数时候的移动方法 unsigned min_op[10] = {0}; //9种操作对应数组 unsigned operations[10][10] = { {0}, // A B C D E F G H I {0, 1, 1, 0, 1, 1, 0, 0, 0, 0}, //op1: ABDE {0, 1, 1, 1, 0, 0, 0, 0, 0, 0}, //op2: ABC {0, 0, 1, 1, 0, 1, 1, 0, 0, 0}, //op3: BCEF {0, 1, 0, 0, 1, 0, 0, 1, 0, 0}, //op4: ADG {0, 0, 1, 0, 1, 1, 1, 0, 1, 0}, //op5: BDEFH {0, 0, 0, 1, 0, 0, 1, 0, 0, 1}, //op6: CFI {0, 0, 0, 0, 1, 1, 0, 1, 1, 0}, //op7: DEGH {0, 0, 0, 0, 0, 0, 0, 1, 1, 1}, //op8: GHI {0, 0, 0, 0, 0, 1, 1, 0, 1, 1} //op9: EFHI }; //移动操作改变数据clocks2，而clocks备份源数据 unsigned clocks2[10] = {0}; //记录每种移动方法操作次数 unsigned op[10] = {0}; //保存枚举移动1,2,3的64种执行情况 unsigned move123[64][3]; unsigned num, divide_num; for (int i = 0; i < 64; i++) { num = i; //将10进制i转换为对应的4进制move123[i] for (int j = 2; j >= 0; j--) { divide_num = (unsigned)num/4; move123[i][j] = num - divide_num*4; num = divide_num; } } for (int i = 1; i <= 9; i++) cin >> clocks[i]; //枚举移动1,2,3的64种执行次数，计算出移动4-9的移动次数，判断是否满足最后条件，记录总次数 for (int i = 0; i < 64; i++) { memcpy(clocks2, clocks, sizeof(unsigned)*10); memset(op, 0, sizeof(unsigned)*10); for (int j = 1; j <=3; j++) { op[j] = move123[i][3-j]; operate(operations, clocks2, j, op[j]); } //移动操作4,5,6的次数分别由时钟A,B,C的状态决定 for (int j = 4; j <=6; j++) { op[j] = (4-clocks2[j-3])%4; operate(operations, clocks2, j, op[j]); } //移动操作7,9的次数分别由时钟D,F的状态决定 op[7] = (4-clocks2[4])%4; operate(operations, clocks2, 7, op[7]); op[9] = (4-clocks2[6])%4; operate(operations, clocks2, 9, op[9]); //判断E是否为0，GHI是否相等 if (clocks2[5] == 0 && clocks2[7] == clocks2[8] && clocks2[8] == clocks2[9]) { op[8] = (4-clocks2[7])%4; tmp = 0; for (int j = 1; j <= 9; j++) tmp += op[j]; if (tmp < min_res) { min_res = tmp; memcpy(min_op, op, sizeof(unsigned)*10); } } } //非零项排序 vector<unsigned> res; for (int i = 1; i <= 9; i++) { if (min_op[i] != 0) { res.push_back(i); min_op[i]--; i--; } } sort(res.begin(), res.end()); for (vector<unsigned>::iterator it = res.begin(); it != res.end(); it++) cout << *it << ' '; cout << "n"; return 0;}
