V. Array Again

বলে রাখি, আমি ইচ্ছে করে এই কোডে একটি bug রেখেছি  আপনারা জাস্ট debug  করে উপলোড দিবেন or for RIGHT answer ক্লিক হেয়ার 

Problem link Click here

Score: 1

CPU: 1s
Memory: 512MB

You will be given a grid of size n*n. Then you will be given n*n numbers of the grid. Now you have to print the numbers of grid in spiral way.

Suppose you have a grid of size 3 * 3 and the given numbers are 1 2 3 4 5 6 7 8 9. Then the number will be in the grid like below.

1 2 3
4 5 6
7 8 9 

Now you have to print numbers as 1 2 3 6 9 8 7 4 5

Input
First you will be given an positive integer T (1< T <= 40), the number of test case. For each test case you will be given n( 0 < n < 50 ), the size of the grid. Then there will be n lines, each containing n numbers. No numbers will be greater than 9.

Output
Output For each test case you have to print “Case X: ”, where X is the number of test case and n*n numbers separated by a single space in spiral order, from top left side of the grid.

Sample

Input	Output
1 3 1 2 3 4 5 6 7 8 9	Case 1: 1 2 3 6 9 8 7 4 5

Solution

#include <stdio.h>
int main()
{
    int row,t,i, col, n, matrix[49][49];
    scanf("%d",&t);
    for(i=1;i<=t;i++){
    scanf("%d", &n);
    for(row = 1; row <= n; row++){
        for(col = 1; col <= n; col++){
            scanf("%d", &matrix[row][col]);
        }
    }
    row = 1;
    printf("Case %d:",i);
    while(n > 0){
        for(col = row; col <= n; col++){
            printf(" %d",matrix[row][col]);
        }
        for(col = row+1; col <= n; col++){
            printf(" %d", matrix[col][n]);
        }
        for(col = n-1; col > row-1; col--){
            printf(" %d", matrix[n][col]);
        }
        for(col = n-1; col > row; col--){
            printf(" %d", matrix[col][row]);
        }
        row++;
        n--;
    }
    printf("\n");
    }
    return 0;
}