A Simple Chess Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2597 Accepted Submission(s): 691 Problem Description There is a n×m board, a chess want to go to the position (n,m) from the pos
题目:给定一个目录,判断该目录的大小,单位为G 思路: 递归拿到目录的子文件,然后取长度,累加 public class FileDemo02 { public static void main(String[] args) { FileDemo02 demo = new FileDemo02(); long size = demo .printDirSize(new File("D:/视频/极客学院知识体系/Java/4.Java语言进阶")); System.out.println
#include<stdio.h> #include<stdlib.h> int data[100][100] = {0}; int Dx[8] = {-1,1,-2,2,-2,2,-1,1}; int Dy[8] = {-2,-2,-1,-1,1,1,2,2}; int row,col; int cun[100][2] = {0}; bool ret = false; void DFS(int x,int y,int step); int main() { int n; //fr
Knight Moves Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8831 Accepted Submission(s): 5202 Problem Description A friend of you is doing research on the Traveling Knight Problem (TKP) wher