1105 Spiral Matrix（25 分）
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 mrows 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：输入一个数N，并且有N个数，给出一个非递增的螺旋矩阵，m是行，n是列，m*n=N，要求m>=n,并且在所有可能的取值中，m-n的值最小。
//1.首先就需要对N进行因式分解，确定m和n。2.其次就放就可以了。
//N的范围没给。 不太明白螺旋矩阵的下标是什么规律。需要先学习一下螺旋矩阵的算法。
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
int func(int N) {
int i = sqrt((double)N);
while(i >= ) {
if(N % i == )//i是这样递减的，那么肯定能够保证是差值最小的了。
return i;
i--;
}
return ;
}
int cmp(int a, int b) {return a > b;}
int main() {
int N, m, n, t = ;
scanf("%d", &N);
n = func(N);//如果是12，那么n=3；
m = N / n;
vector<int> a(N);
for (int i = ; i < N; i++)
scanf("%d", &a[i]);
sort(a.begin(), a.end(), cmp);//从大到小排列。
vector<vector<int> > b(m, vector<int>(n));//m行，这么多列。
int level = m / + m % ;
for (int i = ; i < level; i++) {//i表示是第几圈。
for (int j = i; j <= n - - i && t <= N - ; j++)//t是控制原数组不越界。
b[i][j] = a[t++];//右
for (int j = i + ; j <= m - - i && t <= N - ; j++)
b[j][n - - i] = a[t++];//下
for (int j = n - i - ; j >= i && t <= N - ; j--)
b[m - - i][j] = a[t++];//左
for (int j = m - - i; j >= i + && t <= N - ; j--)
b[j][i] = a[t++];//上
}
for (int i = ; i < m; i++) {
for (int j = ; j < n; j++) {
printf("%d", b[i][j]);
if (j != n - ) printf(" ");
}
printf("\n");
}
return ;
}
//看了这个代码，还是稍微有点不太明白，需要复习，一下。
坐标的起始点和控制范围，
首先是向右的时候 ：第几圈就是那个起始点，那么列的范围就是n-1-i；向下的时候，行的范围就是i+1开始，n-1-i列，这个需要多去实现一下。