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:

#include<cstdio>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = ; //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 = ; i < N; i++){

        scanf("%d",&A[i]);

    }

    if(N == ){

        printf("%d",A[]);

        return ;

    }

    sort(A,A+N,cmp);

    int m = (int)ceil(sqrt(1.0*N));

    while(N % m != ) m++;  //除不整的时候m++

    int n = N / m, i = , j = , now = ;

    int U = , D = m, L = , 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 - ){

            matrix[i][j] = A[now++];

        }

    }

    for(int i = ; i <= m; i++){

        for(int j = ; j <= n; j++){

            printf("%d",matrix[i][j]);

            if(j < n) printf(" "); //j < n,不是j < n - 1

            else printf("\n");

        }

    }

    return ;

}

巴特西

1105 Spiral Matrix（25 分）

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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