littleorange6的博客

UVa 340 #include #define MAX 1020 int main() { int n, i, cas = 1; int code[MAX], guess[MAX], hash1[20], hash2[20]; while(1) { scanf("%d", &n); if(!n) break; f

Uva 10055 #include int main() { unsigned long int a, b, c; while(scanf("%lu%lu", &a, &b) == 2) { c = a > b ? a - b : b - a; printf("%lu\n", c); } return 0; } sum

import java.io.*; import java.util.*; class Node { int val; Node left; Node right; public Node(int val) { this.val = val; } } public class Main { static Integer[] inorder; static Integer[

四分树的一个不一样的思路——单纯的递归实现 #include int board[32][32]; char str[20000]; void giveblack(int length, int left, int up) { int i, j; for(i = up; i < up + length; i++) for(j = left; j < left+

1.用完全二叉树 #include int order[10]; int fulltree[2000]; int ans[10]; char str[2000]; int calculate(char* str, int n) { int i; int p = 0; for(i = 0; i < n; i++) { if( str[order[i

import java.io.File; import java.io.FileNotFoundException; import java.util.LinkedList; import java.util.Scanner; class Stack{ private LinkedList storage = new LinkedList(); public Stack(String str

#include #define HASHSIZE 20000 typedef struct TreeNode { int val; int pos; struct TreeNode* left; struct TreeNode* right; }Node; int arr[HASHSIZE]; int hashmap[HASHSIZE]; int ans[HAS

#include #define N 8 int vis[N][N]; int step[N][N]; int dx[8] = {1,2,2,1,-1,-2,-2,-1}; int dy[8] = {2,1,-1,-2,-2,-1,1,2}; int bfs(int x, int y, int target) { int front = 0, rear = 0, u, d; in

#include #define SIZE 110 int ROW, COL; int dx[8] = {0,1,1,1,0,-1,-1,-1}; int dy[8] = {1,1,0,-1,-1,-1,0,1}; char map[SIZE][SIZE]; int visited[SIZE][SIZE]; void dfs(int row, int col) { if(visited[