PAT甲题题解-1105. Spiral Matrix (25)-（模拟顺时针矩阵）

　　题意：给定N，以及N个数。找出满足m*n=N且m>=n且m-n最小的m、n值，建立大小为m*n矩阵，将N个数从大到下顺时针填入矩阵中。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <cmath>
#define LEFT 1
#define RIGHT 2
#define UP 3
#define DOWN 4
using namespace std;
const int maxn=10000+5;
int a[maxn];
int n,m;
int N;
int main()
{
    int **matrix;
    int tmp;
    scanf("%d",&N);

    tmp=(int)sqrt(N*1.0f+0.5);
    while(N%tmp!=0){
        tmp++;
    }
    m=max(tmp,N/tmp);
    n=min(tmp,N/tmp);

    matrix=new int*[m];
    for(int i=0;i<m;i++)
        matrix[i]=new int[n];
    for(int i=0;i<N;i++)
        scanf("%d",&a[i]);
    sort(a,a+N);

    int p;
    int leftBound=0,rightBound=n-1,upBound=0,downBound=m-1;
    int x=0,y=0;
    int cnt=N;
    if(n>1)
        p=RIGHT;
    else
        p=DOWN;
    while(cnt){
        matrix[y][x]=a[cnt-1];
        if(p==RIGHT){
            x++;
            if(x==rightBound){
                upBound++;
                p=DOWN;
            }
        }
        else if(p==DOWN){
            y++;
            if(y==downBound){
                rightBound--;
                p=LEFT;
            }
        }
        else if(p==LEFT){
            x--;
            if(x==leftBound){
                downBound--;
                p=UP;
            }
        }
        else{
            y--;
            if(y==upBound){
                leftBound++;
                p=RIGHT;
            }
        }
        cnt--;
    }
    for(int i=0;i<m;i++){
        for(int j=0;j<n-1;j++){
            printf("%d ",matrix[i][j]);
        }
        printf("%d",matrix[i][n-1]);
        if(i<m-1)
            printf("\n");
    }
    return 0;
}