1105. Spiral Matrix (25)

旧城等待， 2022-05-28 07:05 126阅读 0赞

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

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题目大意：

代码：

#include<stdio.h>
    #include<algorithm>
    #include<math.h>
    using namespace std;
    bool cmp(int a,int b)
    {
        return a>b;
    }
    int a[100000];
    int b[10000][10000];
    int main()
    {
        int n,m,k,t=0;
        scanf("%d",&n);
    
        k= sqrt((double)n);
        while(k>1) {
            if(n % k == 0)
                break;
            k--;
        }
        m=n/k;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a,a+n,cmp);
        for (int i=0;t<n;i++) {
            for (int j=i;j<k-i&&t<n;j++)
                    b[i][j]=a[t++];
            for (int j=i+1;j<m-1-i&&t<n;j++)
                    b[j][k-1-i]=a[t++];
            for (int j=k-i-1;j>=i &&t<n;j--)
                    b[m-1-i][j]=a[t++];
            for (int j=m-2-i;j>=i+1&&t<n;j--)
                    b[j][i]=a[t++];
        }
            for(int i=0;i<m;i++)
            {
                for(int j=0;j<k;j++)
                {
                    if(j==0)
                    {
                        printf("%d",b[i][j]);
                    }
                    else
                    {
                        printf(" %d",b[i][j]);
                    }
                }
                printf("\n");
            }
    }