PTA Spiral Matrix (25分)

释放无限光明的是人心,制造无边黑暗的也是人心,光明和黑暗交织着,厮杀着,这就是我们为之眷恋又万般无奈的人世间。

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×nmust 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 10​4​​. 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 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include//INT_MAX
//#include
#define PP pair
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fll
#define dinf 1000000000000.0
#define PI 3.1415926
typedef long long ll;
using namespace std;
int const mod=1e9+7;
const int maxn=1000100;
int N,s[10010],mp[10010][10010];
int mi=inf,m,n;
int main(){
    cin>>N;
    for(int i=1;i<=N;i++)
        cin>>s[i];
    sort(s+1,s+1+N);
    for(int i=1;i<=sqrt(N);i++){
        if(N%i==0){
            int ls=i;
            if(abs(ls-N/ls)=1;i--){
        if(d==1){
            x++;
            if(x>m||mp[y][x]!=0){
                d=2;
                x--;
                y++;
            }
        }
        else if(d==2){
            y++;
            if(y>n||mp[y][x]!=0){
                d=3;
                y--;
                x--;
            }
        }
        else if(d==3){
            x--;
            if(x<1||mp[y][x]!=0){
                d=4;
                x++;
                y--;
            }
        }
        else{
            y--;
            if(y<1||mp[y][x]!=0){
                d=1;
                y++;
                x++;
            }
        }
        mp[y][x]=s[i];
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cout<

 

你可能感兴趣的:(C++编程)