PAT_A1105#Spiral Matrix

Source:

PAT A1105 Spiral Matrix (25 分)

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:

 /*
 Data: 2019-06-08 16:12:47
 Problem: PAT_A1105#Spiral Matrix
 AC: 01:05:29

 题目大意：
 序列从大到小，顺时针放入矩阵中
 */
 #include<functional>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 const int M=1e4+;
 int matrix[M], a[M];

 int main()
 {
 #ifdef    ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif

     int N,n,m;
     scanf("%d", &N);
     for(int i=; i<N; i++)
         scanf("%d", &a[i]);
     for(int i=(int)sqrt((double)N); i>=; i--)
     {
         if(N%i == )
         {
             n = i;
             m = N/i;
             break;
         }
     }
     sort(a,a+N,greater<int>());
     fill(matrix,matrix+M,);
     int pt=,c=,r=,R=m,C=n;
     while(pt < N)
     {
         for(int i=c; i<=n && pt<N; i++)
             matrix[C*r-+i-]=a[pt++];
         r++;
         for(int i=r; i<=m && pt<N; i++)
             matrix[C*i-+n-]=a[pt++];
         n--;
         for(int i=n; i>=c && pt<N; i--)
             matrix[C*m-+i-]=a[pt++];
         m--;
         for(int i=m; i>=r && pt<N; i--)
             matrix[C*i-+c-]=a[pt++];
         c++;
     }
     for(int i=; i<=R; i++)
         for(int j=; j<=C; j++)
             printf("%d%c", matrix[C*i-+j-], j==C?'\n':' ');

     return ;
 }