PTA 乙级 1050 螺旋矩阵 (25分) C++

 

 这题真的是想了老半天,也不太会,看了看别人的,学习学习

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<cmath>
 5 using namespace std;
 6 /*用于排序的二维数组*/
 7 int out[10000][100] = { 0 };
 8 /*sort*/
 9 bool cmp(int a, int b) {
10     return a > b;
11 }
12 int main() {
13     int num = 0;
14     int n = 0, m = 0;
15     /*写入二维数组时的递增变量*/
16     int a = 0;
17     /*四边形的每个边的层数*/
18     int lel = 0;
19     cin >> num;
20     vector<int> arr(num);
21     for (int i = 0; i < num; ++i)cin >> arr[i];
22     sort(arr.begin(), arr.end(), cmp);                //输入数据从大到小排序
23     for (n = sqrt((double)num); n >= 1; n--) {        //利用平方根寻找符合条件的m和n
24         if (num % n == 0) {
25             m = num / n;
26             break;
27         }
28     }
29     lel = m / 2 + m % 2;                            //长(正)方形边包含最多数字的个数(以最长的边为基准)
30     for (int i = 0; i < lel; ++i) {
31         for (int j = i; j < n - i && a < num; ++j)out[i][j] = arr[a++];                    //
32         for (int j = i + 1; j < m - i - 1 && a < num; ++j)out[j][n - i - 1] = arr[a++]; //
33         for (int j = n - i - 1; j >= i && a < num; --j)out[m - i - 1][j] = arr[a++];    //
34         for (int j = m - i - 2; j > i && a < num; --j)out[j][i] = arr[a++];                //
35     }
36     for (int i = 0; i < m; ++i) {                    //输出
37         for (int j = 0; j < n; ++j) {
38             if (j == n - 1)cout << out[i][j];
39             else cout << out[i][j] << ' ';
40         }
41         cout << endl;
42     }
43     return 0;
44 }

 

 大佬

posted @ 2020-08-15 16:16  上帝的绵羊  阅读(159)  评论(0编辑  收藏  举报