解题思路:分治思想, 递归实现
注意事项:循环次数是2的n次方
参考代码:
#include<iostream> #include<limits> #include<cstring> #include<cmath> using namespace std; const int N = 1e3 + 10; int a[N][N]; void dfs(int n){ if(n == 0)return; dfs(n - 1); int m = pow(2,n-1); for(int i = 1; i <= m; i ++) for(int j = 1; j <= m; j ++){ a[i][m + j] = a[i][j] + m; a[i + m][j + m] = a[i][j]; a[i + m][j] = a[i][j] + m; } } int main() { int n; cin >> n; a[1][1] = 1; dfs(n); int m = pow(2,n); for(int i = 1; i <= m; i ++){ for(int j = 1; j <= m; j ++){ cout << a[i][j] << ' '; } cout << endl; } return 0; }
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