PAT甲级——1105 Spiral Matrix (螺旋矩阵)

此文同步发布在CSDN：https://blog.csdn.net/weixin_44385565/article/details/90484058

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:

98 95 93
42 37 81
53 20 76
58 60 76

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 = 0,
    ans[10000][10000];
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 = 0; i < N; i++)
        scanf("%d", &v[i]);
    sort(v.begin(), v.end(), cmp);
    n = getNum(N);
    m = N / n;
    int left = 0,
        right = n-1,
        bottom = m-1,
        top = 0;
    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 = 0; i < m; i++){
        for(int j = 0; j < n; j++){
            printf("%d", ans[i][j]);
            if(j < n-1)
                printf(" ");
        }
        printf("
");
    }
    return 0;
 } 
 
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 > 1){
        if(N % n == 0)
            break;
        n--;
    }
    return n;
}
bool cmp(int a, int b){
    return a > b;
}