#include<stdio.h>
#define N 20

void main()
{
    jhs(2,2,1);
jds(2,2,1);

}


void jhs(float x,int n,int i)
......................

#include"stdio.h"
int main(){
printf("hellow,world.\n");
return 0;
}

#include <stdio.h>
#include "sqstack.h"

int main (int argc,char *argv[])
{
SqStack sta;
int score[8]={1,2,3,4,5,6,7,8};
InitStack(&sta);
int i;
for(i=0;i<8;i++)
{
if(!Push(&sta,score[i]))
......................

//好习惯
#include <iostream>
#include <string>

struct Free{
    std::string name;
    int num;
};

void Display(const Free & nb);

int main()
......................

//坏习惯
#include <iostream>
#include <string>

struct free{
    std::string name;
    int num;
};

void display(const free & nb);

int main()
......................

y==fabs(pow(x,2)-7*exp(x))+sqrt(3*pow(x,2)+sin(67.0/180*3.1415972));

本人需要开发一个网页采集软件，采集速度要非常快的，而且采集要求够稳定，有意者麻烦联系我QQ：1134665443

rs[maxAccount].accNumber = accNumber;  
    strcpy(rs[maxAccount].accName, accName);                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    
                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          
    strcpy(rs[maxAccount].accPWD, accPWD);  
    rs[maxAccount].balance = cash;  
    strcpy(rs[maxAccount].flag, "0");

/**
 ============================================================================
 【问题描述】
 有这么一个速配电视节目。N位男士和N位女士要在摄像机前选出他们合适的伴侣。每位女士按照其对
 每位男士作为配偶的偏爱程度给每位男士排名次，每位男士也按照其对每位女士作为配偶的偏爱程度给每位
 女士排名次。这些名次不允许并列。然后每位男士将向心仪的对象求婚，经过"残酷"的竞争之后各自找到适
 合的伴侣。最开始的时候每位男士都还没有被任何一位女士拒绝。求婚环节会经过很多轮进行，每一轮：
 (1) 每位男士在还没有拒绝过自己的女士中选出自己认为最理想的一个，并向她求婚
 (2) 每位女士在所有这一轮中向她求婚的男士中选出自己认为最理想的一个，并不答应，也不拒绝。
 她把其余向她求婚的男士都婉言拒绝掉。经过了若干轮求婚之后，在某一轮，幸运的事情发生了：所有的女
 士都恰好有一个求婚者，所有的男士都找到一个心仪的对象。主持人将继续指出这个配对方式的神奇之处：
 没有任意的两个配对，比方说男士A和女士a配对,男士B和女士b配对，使得在A心目中b较a更理想，而且在
......................

#include<stdio.h>
void main()
{
int a,i;
int A[10];
printf("please input a nomber:\n");
scanf("%d",&a);
for(i=0;i<10;i++)
{
A[i]=a%2;
a=a/2;
}
......................