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
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<math.h>
using namespace std;
int ans[][]; bool cmp(int a,int b)
{
return a > b;
}
int main()
{
int len,tem;
scanf("%d",&len);
vector<int> vv;
for(int i = ;i <= len ;++i)
{
scanf("%d",&tem);
vv.push_back(tem);
}
sort(vv.begin(),vv.end(),cmp);
int n,m;
n = sqrt((double)len);
while(len % n != )
{
--n;
}
m = len / n;
int high = , low = m, left = ,right = n;
int x = ,y = ;
for(int i = ;i < len ;++i)
{
while(i < len && x <= right)
{
ans[y][x] = vv[i];
++i;
++x;
}
--x;
++y;
++high;
while(i < len && y <= low)
{
ans[y][x] = vv[i];
++i;
++y;
}
--y;
--x;
--right;
while(i < len && x >= left)
{
ans[y][x] = vv[i];
++i;
--x;
}
++x;
--y;
--low;
while(i < len && y >= high)
{
ans[y][x] = vv[i];
++i;
--y;
}
++y;
++x;
++left;
--i;
}
for(int i = ;i <= m ;++i)
{
for(int k = ;k <= n ;++k)
{
if(k == ) printf("%d",ans[i][k]);
else printf(" %d",ans[i][k]);
}
printf("\n");
}
return ;
}