1105. Spiral Matrix (25)

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and n columns, where m and n satisfy the following: m*n must be equal to N; m>=n; and m-n is the minimum of all the possible values.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 10⁴. The numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.

Sample Input:

12
37 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 93
42 37 81
53 20 76
58 60 76

一圈圈来，每一圈分四边，如果最内一圈只有一列，四个边只需要两个边。代码中有if语句。
代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iomanip>

using namespace std;
int s[10000],n,a,b,c;
int t[10000][10000];
int main()
{
    cin>>n;
    for(int i = 0;i < n;i ++)
    {
        cin>>s[i];
    }
    a = n;
    while(a && a >= n / a)a --;
    a ++;
    while(n % a)a ++;
    b = n / a;
    sort(s,s + n);
    c = n - 1;
    for(int i = 0;i <= min(a,b) / 2;i ++)
    {
        for(int j = i;j < b - i;j ++)
            t[i][j] = s[c --];
        for(int j = i + 1;j < a - i;j ++)
            t[j][b - 1 - i] = s[c --];
        if(a - 1 - i != i)
        for(int j = b - i - 2;j >= i;j --)
            t[a - 1 - i][j] = s[c --];
        if(b - 1 - i != i)
        for(int j = a - i - 2;j >= i + 1;j --)
            t[j][i] = s[c --];
    }
    for(int i = 0;i < a;i ++)
    {
        cout<<t[i][0];
        for(int j = 1;j < b;j ++)
            cout<<' '<<t[i][j];
        cout<<endl;
    }
}