PAT 甲级 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:

1237 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 9342 37 8153 20 7658 60 76

#include <algorithm>#include <cstdio>#include <iostream>#include <vector>#include <queue>using namespace std;int func(int N) {int i = sqrt((double)N);while (i >= 1) {if (N%i == 0)return i;i--;}return 1;}int cmp(int a, int b) { return a > b; }int main() {int N, m, n, t = 0;scanf("%d", &N);n = func(N);m = N / n;vector<int> a(N);for (int i = 0; i < N; i++)scanf("%d", &a[i]);sort(a.begin(), a.end(), cmp);vector < vector<int> > b(m, vector<int>(n));int level = m / 2 + m % 2;for (int i = 0; i < level; i++) {for (int j = i; j <= n - 1 - i && t <= N - 1; j++)b[i][j] = a[t++];for (int j = i + 1; j <= m - 2 - i && t <= N - 1; j++)b[j][n - 1 - i] = a[t++];for (int j = n - i - 1; j >= i && t <= N - 1; j--)b[m - 1 - i][j] = a[t++];for (int j = m - 2 - i; j >= i + 1 && t <= N - 1; j--)b[j][i] = a[t++];}for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {printf("%d", b[i][j]);if (j != n - 1) printf(" ");}printf("\n");}return 0;}