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.

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

98 95 93
42 37 81
53 20 76
58 60 76
<br />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。<br />代码：

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

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