PAT甲级——A1105 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 Npositive integers to be filled into the spiral matrix. All the numbers are no more than 1. 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

就是一个分块思想

 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 using namespace std;
 int nn, m, n;
 int main()
 {
     cin >> nn;
     vector<int>v(nn);
     for (int i = ; i < nn; ++i)
         cin >> v[i];
     n = floor(sqrt(nn));//取小值
     while (nn%n!=)n--;//找到m,n
     m = nn / n;
     vector<vector<int>>arry(m, vector<int>(n, ));
     sort(v.begin(), v.end(), [](int a, int b) {return a > b; });
     int lm = , ln = ;//左上角
     int rm = m - , rn = n - ;//右下角
     int k = ;//使用数据的下角标
     while (lm <= rm && ln <= rn && k < nn)
     {
         if (lm == rm)//只有一行，则打印
             for (int i = ln; i <= rn; ++i)
                 arry[lm][i] = v[k++];
         else if (ln == rn)//只有一列
             for (int i = lm; i <= rm; ++i)
                 arry[i][ln] = v[k++];
         else
         {
             for (int i = ln; i < rn; ++i)//上行
                 arry[lm][i] = v[k++];
             for(int i=lm;i<rm;++i)//右列
                 arry[i][rn]= v[k++];
             for (int i = rn; i > ln; --i)//下行
                 arry[rm][i] = v[k++];
             for (int i = rm; i > lm; --i)
                 arry[i][ln] = v[k++];
         }
         lm++, ln++;//左上角右下移
         rm--, rn--;//右下角左上移
     }
     for (int i = ; i < m; ++i)
     {
         for (int j = ; j < n; ++j)
             cout << arry[i][j] << (j == n -  ? "" : " ");
         cout << endl;
     }
     return ;
 }