蛇形填数

好吧，我承认题目不难，只是我遗忘的太多了；正在一点点的找回来。这个版本的蛇形填数是从(0,0)开始横着填的，

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=12;
int main()
{int oi[N][N];
    int n;
    while(cin>>n)
    {if(n>N) break;
        memset(oi,0,sizeof(oi));
   int  count=1;
    int i=0,j=0;
while(count<n*n){
    while(j<n-1&&!oi[i][j+1]){oi[i][j++]=count++;}
    while(i<n-1&&!oi[i+1][j]){oi[i++][j]=count++;}
    while(j>=1&&!oi[i][j-1]){oi[i][j--]=count++;}
    while(i>=1&&!oi[i-1][j]){oi[i--][j]=count++;}
  }oi[i][j]=count;
for(i=0;i<n;i++)
        {for(j=0;j<n;j++)
   printf("%3d\t",oi[i][j]);
    cout<<endl;}
    }
}

　　这个是从(0,n-1)开始的

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=12;
int main()
{int oi[N][N];
    int n;
    while(cin>>n)
    {if(n>N) break;
        memset(oi,0,sizeof(oi));
   int  count=1;
    int i=0,j=n-1;
while(count<n*n){
    while(i<n-1&&!oi[i+1][j]){oi[i++][j]=count++;}
        while(j>=1&&!oi[i][j-1]){oi[i][j--]=count++;}
    while(i>=1&&!oi[i-1][j]){oi[i--][j]=count++;}

    while(j<n-1&&!oi[i][j+1]){oi[i][j++]=count++;}
}oi[i][j]=count;
for(i=0;i<n;i++)
        {for(j=0;j<n;j++)
   printf("%3d\t",oi[i][j]);
    cout<<endl;}
    }
}