思路來源：蛇形填數(一)

#include <iostream>
#include <iomanip>
using namespace std;
int a[100][100];
void main(){
    int n,count,x,y;
    cin>>n;
    memset(a,0,sizeof(a));   //初始化為0
    count = a[x=0][y=n-1] = 1;
    while(count<n*n){
        while(x+1<n && !a[x+1][y]) a[++x][y]=++count;
        while

(y-1>=0 &&!a[x][y-1]) a[x][--y]=++count;
        while(x-1>=0 &&!a[x-1][y]) a[--x][y]=++count;
        while(y+1<n &&!a[x][y+1]) a[x][++y] =++count; } for(x=0;x<n;x++){      //輸出
        for(y=0;y<n;y++){
            cout<<setw(5)<<a[x][y];
        }
        cout<<endl;
    }
}

本題程式碼：

#include <stdio.h>
#include <string.h>
#define M 1000
int a[M][M];
int main(){
 int N, n, max, count, x, y;
 scanf("%d", &N);
 while(N--){
  scanf("%d", &n);
  max = (1 + n) * n / 2;
  count = 0;
  a[x = 1][y = 1] = ++count;
  while(count < max){
   while(y < n - x + 1){
    if(a[x][y + 1] == 0) a[x][++y] = ++count;
    else break;
   }//向右
   while(y > 1){
    if(a[x + 1][y - 1] == 0) a[++x][--y] = ++count;
    else break;
   }//向左下
   while(x > 1){
    if(a[x - 1][y] == 0) a[--x][y] = ++count ;
    else break;//向上
   }   }
  for(x = 1; x <= n; ++x){
   for(y = 1; y <= n - x + 1; ++y) printf("%d ", a[x][y]);
   printf("\n");
  }
  if(N) printf("\n");
  memset(a, 0, sizeof(a));
 }
 return 0;
}