pat-a1105. Spiral Matrix (25)

水题。。还调了一段时间，惭愧。。刚开始在while循环里面多加了一个i++。。。

#include<cstdio>
#include<functional> 
#include<algorithm>
#include<cmath>
using namespace std;
int num[1000000];
int ans[10000][10000];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;++i) scanf("%d",num+i);
	sort(num,num+n,greater<int>());
	int x=(int)sqrt(n);
	int y=n/x;
	while(x*y!=n){
		x--;
		y=n/x;
	}
	int i=0;
	int a=1,b=0;
	while(i<n){
		while(b<x&&ans[a][b+1]==0) ans[a][++b]=num[i++];
		while(a<y&&ans[a+1][b]==0) ans[++a][b]=num[i++];
		while(b>1&&ans[a][b-1]==0) ans[a][--b]=num[i++];
		while(a>1&&ans[a-1][b]==0) ans[--a][b]=num[i++];
	}
	for(int i=1;i<=y;++i){
		for(int j=1;j<=x-1;++j){
			printf("%d ",ans[i][j]);
		}
		printf("%d\n",ans[i][x]);
	}
	return 0;
} 

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