Dim rag As Range
For Each rag In Range("数独盘")
rag.ClearContents
rag.Font.Bold = False
Next
End Sub
用如下VBA程序把初盘“粗体化”：Sub 粗体初盘()
Dim rag As Range
For Each rag In Range("数独盘")
If rag.Value = "" Then
rag.Font.Bold = False
Else
rag.Font.Bold = True
End If
Next
End Sub为建“可选数”表，先用如下程序语句清空“可选数”区域：Range("可选数").ClearContents '清空可选数
以下程序段，作为解数独的初始化工作的一部分完成如下工作：在“数独盘”区域仅保留粗体的已知数作为初盘。在“可选数”区域的与“数独盘”已知数对应的单元格，抄录已知数，其他单元格取“123456789”。程序如下：For Each rag In Range("可选数")
If rag.Offset(-10, 0).Font.Bold Then '“数独盘”中用粗体字表示已知数
rag.Value = rag.Offset(-10, 0).Value '把“数独盘”中的已知数抄录到对应的“可选数”区域中的相应位置
Else
rag.Value = "123456789" '“数独盘”中的未知数在“可选数”区域中的相应位置设置成“123456789”
rag.Offset(-10, 0).Value = "" '“数独盘”中的未知数用空格表示
End If
Next
接下来的初始化工作就是建立对应于初盘的可选数表。这部分工作和以后要讲的用“排除法”解数独题有相似之处。这些相似的工作用“处理确定数”子程序来完成。现在在“可选数”区域，寻找长度为1的数作为确定数，把该单元格置为空格，记下该单元格的行号和列号，由“处理确定数”子程序作相应处理。程序如下：For Each rag In Range("可选数")
If Len(rag.Value) = 1 Then
qds = rag.Value '已知数作为“确定数”
rag.Value = "" '对应于“确定数”在“可选数”区域设为空格，表示无别的数可选
i% = rag.Row - Range("可选数").Row + 1 '计算“确定数”在“可选数”区域的行号
j% = rag.Column - Range("可选数").Column + 1 '计算“确定数”在“可选数”区域的列号
Call 处理确定数(i%, j%, qds)
End If
Next
待续
前两天刚写完,还没优化,已运行通过了.晕，一维的好麻烦，这个也是碰巧前两天刚写好的，你看着自己修改下
#include <stdio.h>
typedef struct
{
int line
int row
int num
}Node
int main()
{
/*
int a[9][9]={
{4,0,3,6,0,0,0,0,0},
{0,0,0,0,0,1,0,2,4},
{0,1,0,0,4,0,5,0,0},
{0,0,0,9,0,4,0,6,0},
{3,0,2,0,0,0,4,0,9},
{0,7,4,1,0,3,0,0,0},
{0,0,1,0,9,0,0,4,0},
{2,4,0,3,0,0,0,0,0},
{0,0,0,4,0,8,2,0,7}}
*/
int a[9][9]={
{0,0,0,8,0,0,0,6,0},
{8,7,0,0,0,0,0,0,0},
{2,9,0,0,4,1,0,0,5},
{0,0,5,7,0,0,0,0,9},
{0,2,0,0,0,0,0,1,0},
{9,0,0,0,0,4,3,0,0},
{7,0,0,6,1,0,0,9,8},
{0,0,0,0,0,0,0,5,2},
{0,6,0,0,0,9,0,0,0}}
/*
int a[9][9]={
{0,2,0,0,6,0,0,0,0},
{0,9,0,4,0,5,1,3,0},
{0,0,8,7,0,0,0,0,5},
{6,0,0,3,0,0,4,0,0},
{0,0,0,9,0,6,0,0,0},
{0,0,7,0,0,1,0,0,3},
{4,0,0,0,0,7,3,0,0},
{0,8,5,2,0,4,0,7,0},
{0,0,0,0,9,0,0,1,0}}
*/
/*
int a[9][9]={
{0,0,3,0,2,0,0,0,6},
{0,0,2,0,9,0,0,0,4},
{7,0,0,8,0,0,2,0,3},
{0,8,0,0,7,0,5,0,0},
{0,7,0,1,0,6,0,3,0},
{0,0,0,2,0,0,0,9,0},
{4,0,6,0,0,8,0,0,5},
{6,0,0,0,4,0,3,0,0},
{9,0,0,0,1,0,7,0,0}}
*/
int i,j,n,en,flag,y,k=0,x,qu,p,q
Node b[70]
for(i=0i<9i++)
{
for(j=0j<9j++)
{
if(!a[i][j])
{
b[k].line=i
b[k].row=j
b[k].num=0
k+=1
}
}
}
en=k
/*从b[0]开始试，若b[k].num>9,则k-1,否则k+1*/
for(k=0k<en)
{
++b[k].num
i=b[k].line
j=b[k].row
a[i][j]=b[k].num
n=0
while(n<9&&b[k].num<=9)
{
if(n==i)
{
for(y=0y<9y++)
{
if(y==j)
continue
if(a[n][y]==a[i][j])
flag=1
}
}
else if(n==j)
{
for(y=0y<9y++)
{
if(y==i)
continue
if(a[y][n]==a[i][j])
flag=1
}
}
/*判断同一块中有没有相同值*/
qu=3*(i/3)+j/3
switch(qu)
{
case 0:x=0
y=0
break
case 1:x=0
y=3
break
case 2:x=0
y=6
break
case 3:x=3
y=0
break
case 4:x=3
y=3
break
case 5:x=3
y=6
break
case 6:x=6
y=0
break
case 7:x=6
y=3
break
default :x=6
y=6
break
}
p=x
q=y
for(x<p+3x++)
{
for(y<q+3y++)
{
if(x==i&&y==j)
continue
if(a[x][y]==a[i][j])
{
flag=1
break
}
}
if(flag==1)
break
}
if(flag==1)
{
a[i][j]=++b[k].num
flag=0
n=0
continue
}
n++
}
if(b[k].num>9)
{
a[i][j]=b[k].num=0
k--
if(k<0)
{
printf("error!\r\n")
return -1
}
}
else
k++
}
for(i=0i<9i++)
{
for(j=0j<9j++)
{
printf("%d",a[i][j])
}
printf("\r\n")
}
return 1
}
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