标签：ati can file ini ott ons tput printf upper
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.
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.
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.
12
37 76 20 98 76 42 53 95 60 81 58 93
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("\n"); } } return 0; }
标签：ati can file ini ott ons tput printf upper
原文地址：https://www.cnblogs.com/wanghao-boke/p/9574335.html
