PAT 甲级 1105 Spiral Matrix (25分)（螺旋矩阵，简单模拟）

Posted 2021-03-13 caiyishuai

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

 

题意：

用这个数列的数，形成一个螺旋矩阵（顺时针）从大到小

题解：

1.将数列排序
2.分别填入向右向下向左向上四个方向的数字，一次控制 列坐标增加，行坐标增加，列坐标减少，行坐标减少。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int N;
int a[100005];
int n,m;
int ma[105][105];
int main(){
    memset(a,0,sizeof(a));
    cin>>N;
    for(int i=1;i<=N;i++) cin>>a[i];
    sort(a+1,a+1+N);
    for(int i=1;i<=sqrt(N);i++){
        if(N%i==0){
            m=i;
            n=N/i;
        }
    }
    //cout<<n<<" "<<m<<endl;
    int x=0,y=1;
    int d=1;
    for(int i=N;i>=1;i--){
        if(d==1){
            x++;
            if(x>m||ma[y][x]!=0){
                d=2;
                x--;
                y++;
            }
        }else if(d==2){
            y++;
            if(y>n||ma[y][x]!=0){
                d=3;
                y--;
                x--;
            }
        }else if(d==3){
            x--;
            if(x<1||ma[y][x]!=0){
                d=4;
                x++;
                y--;
            }
        }else{
            y--;
            if(y<1||ma[y][x]!=0){
                d=1;
                y++;
                x++;
            }
        }
        ma[y][x]=a[i];
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cout<<ma[i][j];
            if(j!=m) cout<<" ";
            else cout<<endl;
        }
    }
    return 0;
}