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
AC代码
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
int main(){
int N;
cin>>N;
vector<int> num(N);
for(int i = 0; i < N; i++)
cin>>num[i];
sort(num.begin(), num.end());
int m = (int)sqrt(1.0 * N), n; //矩阵行列数
while(m <= N){
if(N % m == 0){
n = N / m;
if(m >= n) break;
}
m++;
}
int a = m, b = n;
int matrix[m + 1][n + 1];
int i = N - 1, start = 0;
while(m > 0 && n > 0){
for(int j = 0; i >= 0 && j < n; j++)
matrix[start][start + j] = num[i--];
for(int j = 1; i >= 0 && j < m; j++)
matrix[start + j][start + n - 1] = num[i--];
for(int j = n - 2; i >= 0 && j >= 0; j--)
matrix[start + m - 1][start + j] = num[i--];
for(int j = m - 2; i >= 0 && j > 0; j--)
matrix[start + j][start] = num[i--];
start++;
n -= 2;
m -= 2;
}
for(int i = 0; i < a; i++){
for(int j = 0; j < b; j++){
cout<<matrix[i][j];
if(j == b - 1) cout<<endl;
else cout<<" ";
}
}
return 0;
}