44 lines
1.3 KiB
C++
44 lines
1.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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//目标：理解递归，思考怎样把问题分解为更小的同样的问题
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int A[1000][1000];
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int i, j;
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//很显示这是一个递归的函数，要深入理解递归
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void solve(int n) {
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if (n == 1) {
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A[0][0] = 1;//递归边界，标记好第一个元素
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} else {
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int m = n / 2;//划分为四块后，每块的边长为原来的一半
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//求解左上角
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solve(m);
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for (i = m; i < n; i++)//左下角可由左上角对应的每个数加边长得到
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for (j = 0; j < m; j++)
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A[i][j] = A[i - m][j] + m;
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for (i = 0; i < m; i++)//右上角可由左下角复制而得到
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for (j = m; j < n; j++)
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A[i][j] = A[i + m][j - m];
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for (i = m; i < n; i++)//右下角可由左上角复制而得到
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for (j = m; j < n; j++)
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A[i][j] = A[i - m][j - m];
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}
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}
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int main() {
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int k;
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cin >> k;
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//求解边长为n（2的k次幂）的循环日程表
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int n = 1 << k;
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//主函数调用
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solve(n);
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//输出结果
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for (int i = 0; i < n; i++, printf("\n"))
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for (int j = 0; j < n; j++)
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printf("%3d", A[i][j]);
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return 0;
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} |