2.11循环赛日程表

#include <iostream>
#include <cmath>
using namespace std;

void Table(int k, int n, int **a);
void Input(int &k);
void Output(int **a, int n);

int main(){
int k = 0;
Input(k);

int n = 1;
//n = 2^k(k>=1)个选手参加比赛
for (int i = 1; i <= k; ++i)
n *= 2;
//根据n动态分配二维数组a
int **a = new int *[n + 1];
for (int i = 0; i <= n; ++i)
a[i] = new int [n + 1];

Table(k, n, a);

cout << "循环赛事日程表为：" << endl;
Output(a, n);

//释放空间
for (int i = 0; i <= n; ++i)
delete[] a[i];
delete[] a;
return 0;
}

void Input(int &k) {
cout << "请输入k值：";
cin >> k;
}

void Output(int **a, int n) {
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j)
cout << a[i][j] << " ";
cout << endl;
}
}

void Table(int k, int n, int **a) {
for (int i = 1; i <= n; ++i)
a[1][i] = i;//设置日程表第一行
int m = 1;
for (int s = 1; s <= k; ++s) {
n /= 2; //
for (int t = 1; t <= n; ++t) {
for (int i = m + 1; i <= 2 * m; ++i)
for (int j = m + 1; j <= 2 * m; ++j) {
a[i][j + (t - 1) * m * 2] = a[i - m][j - m + (t - 1) * m * 2];
a[i][j - m + (t - 1) * m * 2] = a[i - m][j + (t - 1) * m * 2];
}
}
m *= 2;
}
}
/* m 为了设置每次填充时的起始位置 ；将原问题划分为 K 步骤解决，S 加以控制步骤的顺序；
将每步骤划分为 (n /= 2) 部分，t 控制每部分的初始化顺序; 将（t-1)*m*2 视为每次跨的步距，
每次的起始坐标(i, j)是(i, j)从初始位置(1, 1)，i 下移 m, j 右移 m 后得到的,
若k = 3, 则Table()内 s = {1, 2, 3}; n = {4, 2, 1}; m = {1, 2, 4}; 为对角互相初始化的
推导式提供解释。
*/