Description Fill in 1,2,...,n*n in the n*n square Chen, and it is required to fill in the shape of a snake. For example, when n=4, the formula is:
10 11 12 1
9 16 13 2
8 15 14 3
7 6 5 4
10 11 12 1
9 16 13 2
8 15 14 3
7 6 5 4
- enter
- Directly enter the dimension of Fang Chen, that is, the value of n. (n<=100)
- output
- The output is a snake-shaped square Chen.
- sample input
-
3
- Sample output
-
7 8 1 6 9 2 5 4 3
- source
- Algorithm Classic
- Uploaded by
- CEO
- code show as below:
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=101;
int a[maxn][maxn]={0};
int main()
{
int m, i, x , y ;
while(scanf("%d",&m)!=EOF){
#include<cstring>
using namespace std;
const int maxn=101;
int a[maxn][maxn]={0};
int main()
{
int m, i, x , y ;
while(scanf("%d",&m)!=EOF){
memset(a,0,sizeof(a));
i=a[0][m-1]=1;
x=0;y=m-1;
while(i<m*m){
while(x+1<m && !a[x+1][y]){ //右
a[++x][y]=++i;
//printf("%d %d %d\n",x,y,a[x][y]);
}
while(y-1>=0 && !a[x][y-1]){ //下
a[x][--y]=++i;
//printf("%d %d %d\n",x,y,a[x][y]);
}
while(x-1>=0 && !a[x-1][y]){ //左
a[--x][y]=++i;
//printf("%d %d %d\n",x,y,a[x][y]);
}
while(y+1<m && !a[x][y+1]){ //上
a[x][++y]=++i;
//printf("%d %d %d\n",x,y,a[x][y]);
}
}
for(int i=0;i<m;i++)
{
for(int j=0;j<m;j++)
{
printf("%2d ",a[i][j]);
}
printf("\n");
}
}
i=a[0][m-1]=1;
x=0;y=m-1;
while(i<m*m){
while(x+1<m && !a[x+1][y]){ //右
a[++x][y]=++i;
//printf("%d %d %d\n",x,y,a[x][y]);
}
while(y-1>=0 && !a[x][y-1]){ //下
a[x][--y]=++i;
//printf("%d %d %d\n",x,y,a[x][y]);
}
while(x-1>=0 && !a[x-1][y]){ //左
a[--x][y]=++i;
//printf("%d %d %d\n",x,y,a[x][y]);
}
while(y+1<m && !a[x][y+1]){ //上
a[x][++y]=++i;
//printf("%d %d %d\n",x,y,a[x][y]);
}
}
for(int i=0;i<m;i++)
{
for(int j=0;j<m;j++)
{
printf("%2d ",a[i][j]);
}
printf("\n");
}
}
return 0;
}
}