1105. Spiral Matrix (25)-PAT甲级真题

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:
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 <cmath>#include <vector>#include <cstdio>#include <algorithm>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;}