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 nsatisfy 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

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = 10010; //数字不能太大 
int matrix[maxn][maxn],A[maxn];

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

int main(){
    int N;
    scanf("%d",&N);
    for(int i = 0; i < N; i++){
        scanf("%d",&A[i]);
    }
    if(N == 1){
        printf("%d",A[0]);
        return 0;
    }
    sort(A,A+N,cmp);
    int m = (int)ceil(sqrt(1.0*N));
    while(N % m != 0) m++;  //除不整的时候m++ 
    int n = N / m, i = 1, j = 1, now = 0;
    int U = 1, D = m, L = 1, R = n;
    while(now < N){
        while(now < N && j < R){
            matrix[i][j] = A[now++];
            j++;
        }
        while(now < N && i < D){
            matrix[i][j] = A[now++];
            i++;
        }
        while(now < N && j > L){
            matrix[i][j] = A[now++];
            j--;
        }
        while(now < N && i > U){
            matrix[i][j] = A[now++];
            i--;
        }
        U++,D--,L++,R--;
        i++,j++;
        if(now == N - 1){
            matrix[i][j] = A[now++];
        }
    }
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j++){
            printf("%d",matrix[i][j]);
            if(j < n) printf(" "); //j < n,不是j < n - 1 
            else printf("
");
        }
    }
    return 0;
}