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 ncolumns, 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 93Sample Output:
98 95 93 42 37 81 53 20 76 58 60 76
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;
bool cmp(int a,int b)
{
return a>b;
}
int main()
{
int N;
cin>>N;
int num[N+10];
int ans[N+10][N+10];
for(int i=0;i<N;i++)
{
cin>>num[i];
}
if(N==1)
{
printf("%d",num[0]);
return 0;
}
sort(num,num+N,cmp);
int m=(int)ceil(sqrt(1.0*N));
while(N%m!=0)
{
m++;
}
int n=N/m,i=1,j=1,now=0;
int u=1,d=m,l=1,r=n;
while(now<N)
{
while(now<N && j<r)
{
ans[i][j]=num[now++];
j++;
}
while(now<N && i<d)
{
ans[i][j]=num[now++];
i++;
}
while(now<N && j>l)
{
ans[i][j]=num[now++];
j--;
}
while(now<N && i>u)
{
ans[i][j]=num[now++];
i--;
}
u++,d--,l++,r--;
i++,j++;
if(now==N-1)
{
ans[i][j]=num[now++];
}
}
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
printf("%d",ans[i][j]);
if(j<n)
{
printf(" ");
}
else
{
printf("\n");
}
}
}
return 0;
}