PAT 1105. Spiral Matrix （螺旋输出）

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 hasm rows and n columns, where m andn 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 10⁴. The numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each containsn numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.

Sample Input:

1237 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 9342 37 8153 20 7658 60 76

给你一段降序的序列，让你螺旋输出该序列

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int NN = 1e4+100;int vis[NN][NN],a[NN][NN],b[NN]; int main(){ios::sync_with_stdio(false);int N,i,j,n,m,k,flag;cin>>N;for(i=1;i<=N;i++) cin>>b[i];sort(b+1,b+1+N,greater<int>());for(i=1;i<=sqrt(N);i++) {if(N%i==0) m=i,n=N/i;    }   for(i=1;i<=n;i++) vis[i][m+1]=1;for(i=1;i<=m;i++) vis[n+1][i]=1;for(i=1;i<=n;i++) vis[i][0]=1;k=0,flag=1,i=1,j=1;while(1) {a[i][j]=b[++k];if(k==N) break;vis[i][j]=1;if(flag==1) {if(vis[i][j+1]==0) j++;else flag=2,i++;}else if(flag==2) {if(vis[i+1][j]==0) i++;else flag=3,j--;}else if(flag==3) {if(vis[i][j-1]==0) j--;else i--,flag=4;}else {if(vis[i-1][j]==0) i--;else j++,flag=1;}}for(i=1;i<=n;i++) {for(j=1;j<=m;j++) {if(j!=1) cout<<" ";cout<<a[i][j];}cout<<endl;}return 0;}