PAT A1105 Spiral Matrix (25分)甲级题解

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 10​4​​. 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

思路：

本题是一道模拟题，就按照题目要求的思路进行模拟即可，将序列降序排序后，螺旋式地填充到一个结果数组中，按照最终位置来存，本题有2个细节点：
1.当输入数据只有一个数据的时候需要单独判断输出
2.当矩阵是个N*N的矩阵的时候，此时矩阵螺旋输出后最里面一圈会出现只有一个数据的情况，如果不单独处理会出现死循环，只要判断是最后一个数据的时候将它单独处理即可。

代码实现

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
int ans[110][110];

int main() {
	int num,n,m,sqr,cnt=0;
	cin>>num;
	vector<int> sq(num+1);
	sqr=sqrt(1.0*num);
	while(num%sqr!=0) sqr++;
	m=max(num/sqr,sqr);
	n=min(num/sqr,sqr);
	for(int i=0; i<num; i++) scanf("%d",&sq[i]);
	if(num==1) {
		printf("%d",sq[0]);
		return 0;
	}
	sort(sq.begin(),sq.end(),greater<int> {});
	int row=1,col=1,L=1,UP=1,tn=n,tm=m;
	while(cnt<num) {
		while(cnt<num&&col<n)  //向右走
			ans[row][col++]=sq[cnt++];
		while(cnt<num&&row<m) //向下走
			ans[row++][col]=sq[cnt++];
		while(cnt<num&&col>L) //向左走
			ans[row][col--]=sq[cnt++];
		while(cnt<num&&row>UP)//向上走
			ans[row--][col]=sq[cnt++];
		m--;
		n--;//缩小边界 
		UP++;
		L++;//缩小边界 
		col++;//从第二圈缩小矩阵的第一个元素开始 
		row++;
		if(cnt==num-1)
			ans[row][col]=sq[cnt++];
	}
	for(int i=1; i<=tm; i++) {
		for(int j=1; j<=tn; j++) {
			printf("%d",ans[i][j]);
			if(j<tn) printf(" ");
			else printf("\n");
		}
	}
	return 0;
}