Source:
Description:
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×nmust 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 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
Keys:
- 简单模拟
Attention:
- 基本思路就是把序列按照从大到小的顺序，依次填入矩阵中；
- 直接开1e4的矩阵会超出内存，而且样例中存在m较大,n较小的情况，因此用一维数组代替二维数组
- 循环内的四个for循环需要判断pt<N
Code:
1 /*
2 Data: 2019-06-08 16:12:47
3 Problem: PAT_A1105#Spiral Matrix
4 AC: 01:05:29
5
6 题目大意：
7 序列从大到小，顺时针放入矩阵中
8 */
9 #include<functional>
10 #include<cstdio>
11 #include<cmath>
12 #include<algorithm>
13 using namespace std;
14 const int M=1e4+10;
15 int matrix[M], a[M];
16
17 int main()
18 {
19 #ifdef ONLINE_JUDGE
20 #else
21 freopen("Test.txt", "r", stdin);
22 #endif
23
24 int N,n,m;
25 scanf("%d", &N);
26 for(int i=0; i<N; i++)
27 scanf("%d", &a[i]);
28 for(int i=(int)sqrt((double)N); i>=1; i--)
29 {
30 if(N%i == 0)
31 {
32 n = i;
33 m = N/i;
34 break;
35 }
36 }
37 sort(a,a+N,greater<int>());
38 fill(matrix,matrix+M,0);
39 int pt=0,c=1,r=1,R=m,C=n;
40 while(pt < N)
41 {
42 for(int i=c; i<=n && pt<N; i++)
43 matrix[C*r-2+i-1]=a[pt++];
44 r++;
45 for(int i=r; i<=m && pt<N; i++)
46 matrix[C*i-2+n-1]=a[pt++];
47 n--;
48 for(int i=n; i>=c && pt<N; i--)
49 matrix[C*m-2+i-1]=a[pt++];
50 m--;
51 for(int i=m; i>=r && pt<N; i--)
52 matrix[C*i-2+c-1]=a[pt++];
53 c++;
54 }
55 for(int i=1; i<=R; i++)
56 for(int j=1; j<=C; j++)
57 printf("%d%c", matrix[C*i-2+j-1], j==C?‘\n‘:‘ ‘);
58
59
60 return 0;
61 }
原文地址：https://www.cnblogs.com/blue-lin/p/10991319.html