PAT甲级——1105 Spiral Matrix (螺旋矩阵)

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1105 Spiral Matrix (25 分)

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 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:

题目大意：将N个数字降序顺时针螺旋放入一个矩阵，然后输出矩阵。矩阵的行列数分别为m、n，要求m*n=N 且 m≥n、m-n最小。

思路：n取N的平方根，若N能被n整除，则符合条件，若不能，则n--继续寻找。螺旋放入矩阵有四个边界：left、right、top、bottom；写四个函数在边界之内按照四个方向往矩阵放入数据，进入循环：1、从左往右，到达右边界 right 的时候，top++（上边界往下压一层）；2、从上到下，到达底部，right++；3、从右往左，到达左边界，bottom++；4、从下往上，到达上边界，left++；5、若数据放完了则退出循环。

 #include <iostream>

 #include <vector>

 #include <algorithm>

 #include <cmath>

 using namespace std;

 vector <int> v;

 int N,

     index = ,

     ans[][];

 int getNum(int N);

 bool cmp(int a, int b);

 void leftToRight(int &left, int &right, int &top, int &bottom);

 void topToBottom(int &left, int &right, int &top, int &bottom);

 void rightToLeft(int &left, int &right, int &top, int &bottom);

 void bottomToTop(int &left, int &right, int &top, int &bottom);

 int main()

 {

     int m, n;

     scanf("%d", &N);

     v.resize(N);

     for(int i = ; i < N; i++)

         scanf("%d", &v[i]);

     sort(v.begin(), v.end(), cmp);

     n = getNum(N);

     m = N / n;

     int left = ,

         right = n-,

         bottom = m-,

         top = ;

     while(index < N){

         leftToRight(left, right, top, bottom);

         topToBottom(left, right, top, bottom);

         rightToLeft(left, right, top, bottom);

         bottomToTop(left, right, top, bottom);

     }

     for(int i = ; i < m; i++){

         for(int j = ; j < n; j++){

             printf("%d", ans[i][j]);

             if(j < n-)

                 printf(" ");

         }

         printf("\n");

     }

     return ;

  } 

 void bottomToTop(int &left, int &right, int &top, int &bottom){

     if(top > bottom || index >= N)

         return;

     for(int i = bottom; i>= top; i--){

         ans[i][left] = v[index];

         index++;

     }

     left++;

 }

 void rightToLeft(int &left, int &right, int &top, int &bottom){

     if(left > right || index >= N)

         return;

     for(int i = right; i >= left; i--){

         ans[bottom][i] = v[index];

         index++;

     }

     bottom--;

 }

 void topToBottom(int &left, int &right, int &top, int &bottom){

     if(top > bottom || index >= N)

         return;

     for(int i = top; i <= bottom; i++){

         ans[i][right] = v[index];

         index++;

     }

     right--;

 }

 void leftToRight(int &left, int &right, int &top, int &bottom){

     if(left > right || index >= N)

         return;

     for(int i = left; i <= right; i++){

         ans[top][i] = v[index];

         index++;

     }

     top++;

 }

 int getNum(int N){

     int n = sqrt(N);

     while(n > ){

         if(N % n == )

             break;

         n--;

     }

     return n;

 }

 bool cmp(int a, int b){

     return a > b;

 }

巴特西

PAT甲级——1105 Spiral Matrix (螺旋矩阵)

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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