PAT 1105 Spiral Matrix（25） 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×n must be equal to N; m≥n; and m−n is the minimum of all the possible values.
输入格式： 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.
输出格式： 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.
输入样例：
代码语言：javascript复制12
37 76 20 98 76 42 53 95 60 81 58 93输出样例：
代码语言：javascript复制98 95 93
42 37 81
53 20 76
58 60 76题目分析：模拟题。
AC代码：
代码语言：javascript复制#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
vector<int> v;
int main(void){
int k, tmp;
scanf("%d", &k);
for(int i=0; i<k; i){
scanf("%d", &tmp);
v.push_back(tmp);
}
sort(v.begin(), v.end(), [](int a, int b){return a>b;});
int n = floor(sqrt(1.0*k));
while(k % n!=0)n--;
int m = k / n;
int arr[m][n];
bool st[m][n];
fill(st[0],st[0] n*m,false);
fill(arr[0],arr[0] n*m,0);
int idx = 0;
int dx[4] = {-1,0,1,0},dy[4] = {0,1,0,-1};
int x=0,y=0,d=1;
for(int i=0; i<n*m; i){
arr[x][y] = v[idx ];
st[x][y] = true;
int a=x dx[d], b=y dy[d];
if(a<0 || a>=m || b<0 || b>=n || st[a][b]){
d = (d 1)%4;
a = x dx[d],b=y dy[d];
}
x=a,y=b;
}
for(int i=0; i<m; i){
for(int j=0; j<n; j){
printf("%d",arr[i][j]);
if(j!=n-1)printf(" ");
}
printf("n");
}
return 0;
}
