汪少升的博客

原创 Why Emacs

因为 Emacs 很酷。

原创 PAT 1026 Table tennis

使用了３个优先队列和两个普通队列。

原创 PAT 1086

使用栈进行中序遍历模拟建立树。

原创 PAT1025 反转链表

原创 PAT 1025 反转链表

#include #include #define MAXNODE 100001struct Node{ int data; char addr[8]; char next[8];}node[MAXNODE];struct Node *pnode[MAXNODE];int main(void){ char ad[8]; int nnode, k; in

原创 安装windows操作系统基本方法

安装操作系统

原创 指向函数的指针--回调的基础

花了半天的时间研究回调函数，网上有些写得不对，所以记下备忘。话说回来，，很久没有更新博客了，很多事貌似遗忘了。#includeusing namespace std;class callBackTest;typedef void (callBackTest::*classfunp)(int a,int b,int &c);

原创 PAT 1014 Waiting in Line

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:The s

原创 PAT 1004 Counting Leaves

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.InputEach input file contains one test case. Each case starts with a line

原创 PAT 1050 String Subtraction

Given two strings S1 and S2, S = S1 - S2 is defined to be the remaining string after taking all the characters in S2 from S1. Your task is simply to calculate S1 - S2 for any given strings. However,

原创 排序及函数对象的基本用法

1.bool operator()(para...)2.sort(para1,para2,para3)

转载 java awt 页面布局

import java.awt.*;import java.awt.event.*;import java.applet.*;//public class CardLayoutDemo extends Applet implements ItemListener{ String msg=""; Checkbox winXP,win7,solaris,mac; //Panel o

原创 pat1080 graduate admission

#include#include#include#includeusing namespace std;#define N 40001#define M 101#define K 6struct Applicant{ int id,ge,gi; double fg; int rank; int prefer[K];};struct School{ in

转载 My First C++ Program in Professional C++

//employee.h#ifndef EMPLOYEE_H#define EMPLOYEE_H#include#includenamespace Records{ int const kDefaultStartingSalary=30000; class Employee { public: Employee(); void promote(int inRais

原创 pat4-06 Data_sturcture 搜索树判断

#includeusing namespace std;struct Node{ int data; Node *left,*right;};bool LessTree(int *a,int start,int end,Node *&tree){ if(start==end) { tree=new Node; tree->data=a[start]; tree-

原创 pat Data structure --4-05家谱处理 数据结构实验项目

without tree,we can still solve this problem.the key point is how to

原创 Review pat1055 compare cin with scanf,etc..

Now that we prefer to ,why we use in some acm

原创 pat 1055 how to improve the performance

#include#include#include#includeusing namespace std;struct Person{ char name[9]; int age; int worth;};struct Comp{ bool operator()(Person const &per_a,Person const &per_b) { if(per_a

原创 windows 消息队列

#include#include#includeusing namespace std;struct Message{ string msg; int priority;};struct Cmp{ bool operator()(Message const &msg_a,Message const &msg_b) { return msg_a.priority>msg

原创 并查集的基本操作

#includetypedef int T;T const &Find(T const &element,T *father){ if(father[element]<0) { return element; } else { return father[element]=Find(father[element],father); }}T const &Union(

原创 pat Data_Structure 前缀表达式的计算--堆栈法

well，联合的标记问题。。。。怎么弄？#include#includeusing namespace std;struct element{ char op; double nu;};int main(){ stack st; char s[61]; gets(s); double tmp; for(int i=0;s[i]!='\0';++i)

原创 pat Data_stucture 表达式转换-栈的应用

#include#includeusing namespace std;bool p[5][5]={ 1,1,0,0,1, 1,1,0,0,1, 1,1,1,1,1, 1,1,1,1,1, 0,0,0,0,1};int con(char ch){ switch(ch) { case'+': return 0; case'-': return 1; cas

原创 pat Data Stucture 海盗分赃

#include#includeusing namespace std;int Median(int a[],int start,int end){ sort(a+start,a+end); return a[(end-start-1)/2+start];}int main(){ int d,p; cin>>d>>p; int a[p],b[p]; a[p-2]=0;

原创 pat 1079 another solution

this solution shows well in time and space complexity,but it exceeds limited time at last

原创 pat 1079

the key point is how to present a tree....data structuretime and space complexity

原创 pat1029 merge

i want to try list first,but the memory exceeds for three cases.

原创 pat 1078 quadratic probing

#include#include#includeusing namespace std;bool isprime(const int &a){ if(a==1) return false; if(a==2) return true; int len=(int)sqrt((double)a); for(int i=2;i<=len;i++){ if(a%i==0) retu

原创 pat1077

#include#includeusing namespace std;int main(){// freopen("C:\\Users\\Frank Wang\\Desktop\\in.txt","r",stdin); stack s1,s2,s3; int n; scanf("%d",&n); getchar(); char c; while((c=getchar())!=

原创 pat 1017

#include#include#includeusing namespace std;struct client{ double arrive_time; double deal_time; bool operator<(const client &b) const{ return arrive_time<b.arrive_time; }};int main(){ fr