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.

本题要求将给定的 N 个正整数按非递增的顺序，填入“螺旋矩阵”。所谓“螺旋矩阵”，是指从左上角第 1 个格子开始，按顺时针螺旋方向填充。要求矩阵的规模为 m 行 n 列，满足条件：m×n 等于 N；m≥n；且 m−n 取所有可能值中的最小值。

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.

输入在第 1 行中给出一个正整数 N，第 2 行给出 N 个待填充的正整数。所有数字不超过 10​^4​​，相邻数字以空格分隔。

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.

输出螺旋矩阵。每行 n 个数字，共 m 行。相邻数字以 1 个空格分隔，行末不得有多余空格。

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

解题思路

1. 读入待填入数据并计算方阵大小；
2. 填充方阵边界；
3. 模拟螺旋填数过程并标号；
4. 输出结果并返回零值。

代码

#include<cstdio>
#include<cmath>
#include<algorithm>
#define maxn 10010
using namespace std;

struct Node{
    int seq;
    int vivid;
}G[maxn][maxn];

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

int num[maxn];
int N;
int m,n;

void Init(){
    int i;
    scanf("%d",&N);
    n=(int)sqrt(1.0*N);
    for(i=0;i<N;i++){
        scanf("%d",&num[i]);
    }
    sort(num,num+N,cmp);
    while(1){
        if(N%n==0){
            m=N/n;
            break;
        }
        n--;
    }
    for(i=0;i<=m;i++){
        G[i][n].vivid=1;
    }
    for(i=0;i<=n;i++){
        G[m][i].vivid=1;
    }
}

void Op(){
    int k=0;
    int i=0,j=0;
    while(k<N){
        while(G[i][j].vivid==0){
            //右移
            G[i][j].vivid=1;
            G[i][j].seq=k++;
            j++;
        };
        //撞墙退出；
        i++,j--;
        while(G[i][j].vivid==0){
            //下移
            G[i][j].vivid=1;
            G[i][j].seq=k++;
            i++;
        };
        //撞墙退出
        i--,j--;
        while(j>=0&&G[i][j].vivid==0){
            //左移
            G[i][j].vivid=1;
            G[i][j].seq=k++;
            j--;
        };
        //撞墙或出界退出
        i--,j++;
        while(i>=0&&G[i][j].vivid==0){
            //上移
            G[i][j].vivid=1;
            G[i][j].seq=k++;
            i--;
        };
        //撞墙或出界退出
        i++,j++;
    }
}

void Print(){
    int i,j;
    for(i=0;i<m;i++){
        for(j=0;j<n;j++){
            printf("%d",num[G[i][j].seq]);
            if(j<n-1){
                printf(" ");
            }else{
                printf("\n");
            }
        }
    }
}

int main(){
    Init();
    Op();
    Print();
    return 0;
}

运行结果