Question

如何用C解9*9数独，求代码，高手勿喷。

要用C写，而不是C++输入如输出如C++参考代码//weizengke#include <iostream.h>#include<time.h>static int pu[9][9];int isvalid(const int i, const int j)//验证函数当期i,j坐标是否符合游戏规则，不重复{const int n = pu[i][j];const static int query[] = {0, 0, 0, 3, 3, 3, 6, 6, 6};int t, u;for (t = 0; t < 9; t++)if (t != i && pu[t][j] == n || t != j && pu[i][t] == n)//0-9的数字，每行每列都不能重复return 0;for (t = query[i]; t < query[i] + 3; t++) //9个宫的3×3里也不能重复for (u = query[j]; u < query[j] + 3; u++)if ((t != i || u != j) && pu[t][u] == n)return 0; return 1;}void output(void)//输入函数{static int n;cout << "Solution " << ++n << ":" <<endl;for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++)cout<< pu[i][j] << " ";cout << endl;}cout << endl;}void Try(const int n)//核心函数，回溯算法{if (n == 81) {//是否已经是最后一个格子output();return;}const int i = n / 9, j = n % 9;if (pu[i][j] != 0) {//如果当前格子不需要填数字，就跳到下一个格子Try(n + 1);return;}for (int k = 0; k < 9; k++) {pu[i][j]++;//当前格子进行尝试所有解if (isvalid(i, j))Try(n + 1);//验证通过，就继续下一个}pu[i][j] = 0; //如果上面的单元无解，就回溯}int main(void){int i,j;for (i=0;i<9;i++)for (j=0;j<9;j++)cin>>pu[i][j];long start=clock();Try(0);long end=clock();cout<<"计算一共花了"<<(double)(end-start)<<"毫秒"<<endl;return 0;}

Answer 1

int i, j, n=3; int col = 1, line = 0; a[line][col] = 1"%d ", a[i][j]); printf("\\n"); } return 0; } 我有解