PAT 1105 Spiral Matrix[模拟][螺旋矩阵][难]

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 mrows 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:

题目大意：输入一个数N，并且有N个数，给出一个非递增的螺旋矩阵，m是行，n是列，m*n=N，要求m>=n,并且在所有可能的取值中，m-n的值最小。

//1.首先就需要对N进行因式分解，确定m和n。2.其次就放就可以了。

//N的范围没给。不太明白螺旋矩阵的下标是什么规律。需要先学习一下螺旋矩阵的算法。

#include <iostream>

#include <cmath>

#include <vector>

#include <algorithm>

using namespace std;

int func(int N) {

    int i = sqrt((double)N);

    while(i >= ) {

        if(N % i == )//i是这样递减的，那么肯定能够保证是差值最小的了。

            return i;

        i--;

    }

    return ;

}

int cmp(int a, int b) {return a > b;}

int main() {

    int N, m, n, t = ;

    scanf("%d", &N);

    n = func(N);//如果是12，那么n=3；

    m = N / n;

    vector<int> a(N);

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

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

    sort(a.begin(), a.end(), cmp);//从大到小排列。

    vector<vector<int> > b(m, vector<int>(n));//m行，这么多列。

    int level = m /  + m % ;

    for (int i = ; i < level; i++) {//i表示是第几圈。

        for (int j = i; j <= n -  - i && t <= N - ; j++)//t是控制原数组不越界。

                b[i][j] = a[t++];//右

        for (int j = i + ; j <= m -  - i && t <= N - ; j++)

                b[j][n -  - i] = a[t++];//下

        for (int j = n - i - ; j >= i && t <= N - ; j--)

                b[m -  - i][j] = a[t++];//左

        for (int j = m -  - i; j >= i +  && t <= N - ; j--)

                b[j][i] = a[t++];//上

    }

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

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

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

            if (j != n - ) printf(" ");

        }

        printf("\n");

    }

    return ;

}

//看了这个代码，还是稍微有点不太明白，需要复习，一下。

坐标的起始点和控制范围，

首先是向右的时候：第几圈就是那个起始点，那么列的范围就是n-1-i；向下的时候，行的范围就是i+1开始，n-1-i列，这个需要多去实现一下。