Pat(A) 1105. Spiral Matrix (25)

原题目：

原题链接：https://www.patest.cn/contests/pat-a-practise/1105

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.

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 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

题目大意

给N个数字，建立一个m行n列矩阵，要求m>=n，且m-n最小。

并将N个数字由大到小排序后由（0,0）按顺时针方向填入矩阵内。

解题报告

对N个数字排序，再建立m*n矩阵。

代码

#include "iostream"
#include "math.h"
#include "algorithm"
#include "vector"
using namespace std;

int N;
vector<vector<int>> G;
vector<int> data;
int m,n;

bool cmp(int a,int b){
return a>b;
}

void init(){
cin>>N;
int x;
for(int i = 0; i < N; i++){
cin>>x;
data.push_back(x);
}
}

void cal(){
sort(data.begin(),data.end(),cmp);
m = sqrt(N * 1.0);
while(N % m){
m ++;
}
m = max(m,N/m);
n = N/m;
G.resize(m);
for(int i = 0; i < m; i ++)
G[i].resize(n);
}

void build_G(){
int k = 0;
int mm = m - 1,nn = n;
int i = 0,j = 0;
while(k<N){
if(nn){
for(int t = 0; t < nn; t++){
G[i][j] = k;
k ++;
j ++;
}
j --;
i ++;
nn--;
}
if(k >= N)
break;
if(mm){
for(int t = 0; t < mm; t++){
G[i][j] = k;
k ++;
i ++;
}
i --;
j --;
mm--;
}
if(k >= N)
break;
if(nn){
for(int t = 0; t < nn; t++){
G[i][j] = k;
k ++;
j --;
}
j ++;
i --;
nn--;
}
if(k >= N)
break;
if(mm){
for(int t = 0; t < mm; t++){
G[i][j] = k;
k ++;
i --;
}
i ++;
j ++;
mm--;
}
}
}

void print_GData(){
for(int i =0;i<m;i++){
cout<<data[G[i][0]];
for(int j=1;j<n;j++)
cout<<" "<<data[G[i][j]];
cout<<endl;
}
}

int main(){
init();
if(N == 0){
cout<<data[0]<<endl;
return 0;
}
cal();
print_GData();
system("pause");
}