用C语言写的解数独的程序。在linux下测试成功运行。

效果如图：

这是带解的数独，需要填写的部分用数字0代替。

这是程序运行后的效果图。看看，数独已经搞定啦～～～

程序源码如下：

#include <stdio.h>
#include <stdlib.h>#define SIZE 9
#define get_low_bit(x) ((~x&(x-1))+1)struct{int left;char num;	char try;
}board[SIZE][SIZE];int bit2num(int bit)
{switch(bit){case 1:case 2:return bit;	case 4:return 3;case 8:return 4;case 16:return 5;case 32:return 6;	case 64:		return 7;	case 128:return 8;	case 256:return 9;}	
}void printf_res()
{int i, j, k;	for(i=0; i<SIZE; i++){if(i%3==0)	{for(j=0; j<SIZE*2+4; j++)putchar('-');putchar('\n');}		for(j=0; j<SIZE; j++){if(j%3==0)putchar('|');if(board[i][j].num > 0)printf("\033[0;31m%2d\033[0m", board[i][j].num);elseprintf("%2d", board[i][j].try);}	printf("|\n");}for(i=0; i<SIZE*2+4; i++)putchar('-');putchar('\n');
}void sub(int i, int j, int bit)
{int k, m;	for(k=0; k<SIZE; k++){board[k][j].left &= ~bit;board[i][k].left &= ~bit;}		for(k=i/3*3; k<(i/3+1)*3; k++)for(m=j/3*3; m<(j/3+1)*3; m++)board[k][m].left &= ~bit;	
}void init()
{int i, j;	for(i=0; i<SIZE; i++)for(j=0; j<SIZE; j++)if(board[i][j].num > 0)sub(i, j, 1<<(board[i][j].num-1));else if(board[i][j].try > 0)sub(i, j, 1<<(board[i][j].try-1));
}void add(int i, int j, int bit)
{int k, m;for(k=0; k<SIZE; k++){board[k][j].left |= bit;board[i][k].left |= bit;}for(k=i/3*3; k<(i/3+1)*3; k++)for(m=j/3*3; m<(j/3+1)*3; m++)board[k][m].left |= bit;
}void solve(int pos)
{int i=pos/SIZE;	int j=pos%SIZE;	int bit, left;if(pos == SIZE*SIZE){printf_res();exit(0);		}if(board[i][j].num > 0)solve(pos+1);	elsefor(left=board[i][j].left; left; left&=(left-1)){bit = get_low_bit(left);sub(i, j, bit);board[i][j].try = bit2num(bit);solve(pos+1);add(i, j, bit);board[i][j].try=0;init();		}		
}int main()
{int i, j, c;for(i=0; i<SIZE; i++)for(j=0; j<SIZE; j++){while((c=getchar())<'0' || c>'9');board[i][j].num = c-'0';board[i][j].try = 0;board[i][j].left = 0x0001FF;		}				init();solve(0);return 0;
}