一些笔试的代码

1、非递归求最小公倍数和最大公约数

 

#include<stdio.h>
void main()
{
int a,b,num1,num2,temp;
printf("please input num1 and num2 \n");
scanf("%d%d",&num1,&num2);
if(num1 > num2)
{
a = num1;
b = num2;
}
else
{
a = num2;
b = num1;
}

while(b > 0)
{
temp = a % b;
a = b;
b = temp;
}
printf("最大公约数是%d\n最小公倍数是%d\n",a,(num1 * num2) / a);
}

 

 

2、递归求最大公约数

 

//用递归求最大公约数
#include<stdio.h>
int gcd(int m,int n);
int main()
{
  int m,n;
 printf("Input m,n:\n");
  scanf("%d%d",&m,&n);
  printf("%d\n",gcd(m,n));
}
int gcd(int m,int n)
{
 if(m>n)//大于和小于只要"<"或">"就够了，不需要两个
  return gcd(m-n,n);
  else if(m<n)
  return gcd(m,n-m);
 else if(m==n)
  return m;
}

 

3、字符串单词倒置题

 

//字符串转置
#include<string.h>
#include<stdio.h>
void Reversion(char *str)//方法一，利用string.h库函数
{
    int n=strlen(str)-1;
    //char *temp=(char*)malloc(n+1);
    char temp[30]="";
    while(n>0)
    {
        if((str[n]!=' ')&&(str[n-1]==' '))
        {
            strcat(temp,str+n-1);
            str[n]='\0';
        }
        n--;
    }
    strcat(temp,str+n);
    printf("%s\n",temp);
}
void turn(char *str)//方法二，利用循环
{
    char temp;
    int j=strlen(str)-1,i=0,begin,end;
    while(j>i)
    {
        temp=str[i];
        str[i]=str[j];
        str[j]=temp;
        j--;
        i++;
    }
    printf("%s\n",str);
    i=0;
    while(str[i])
    {
        if((str[i]!=' '))
		{
			begin=i;
			while(str[i]&&str[i]!=' ')
				i++;
			end=i-1;
		}
		while(end>begin)
		{
			temp=str[begin];
			str[begin]=str[end];
			str[end]=temp;
			end--;
			begin++;
		}
        i++;
    }
    printf("%s\n",str);
}
void main()
{
    char str[30]="hello world future";
    Reversion(str);
    char str1[]="ni hao a";
    turn(str1);
}

 

 

4、判断低地址还是高地址优先

 

#include<stdlib.h>
#include<stdio.h>
void main()
{
    int a=10;
    short b;
    memcpy(&b,&a,2);//将a的低两字节赋值给b
    printf("%d\n",b);
}

 

 

5、字符串翻转

 

#include <stdio.h>
#include <string.h>
void rotate(char *start, char *end)
{
    while(start != NULL && end !=NULL && start<end)
    {
        char temp=*start;
        *start=*end;
        *end=temp;
        start++;
        end--;
    }
}

void leftrotate(char *p,int m)
{
    if(p==NULL)
        return ;
    int len=strlen(p);
    if(m>0&&m<=len)
    {
        char *xfirst,*xend;
        char *yfirst,*yend;
        xfirst=p;
        xend=p+m-1;
        yfirst=p+m;
        yend=p+len-1;
        rotate(xfirst,xend);
        rotate(yfirst,yend);
        rotate(p,p+len-1);
    }
}

int main(void)
{
    char str[]="abcdefghij";
    leftrotate(str,3);
    printf("%s\n",str);
    return 0;
}

 

6、判断系统大端小端存储

 

#include<stdio.h>
union s{
int i;
char ch;
}c;
int check()
{
c.i=1;
return (c.ch);
}
void main()
{
if (check())
{
printf("little\n");
}
else
printf("big\n");
}

 

7、一道求概率的问题即一个边长为10的正方形和一个半径为10的圆重叠的部分答案为25π左右

 

#include<stdio.h>
#include<time.h>
void main()
{
    int count=0,i=100;
    srand(time(0));
    while(i>0)
    {
        int a=rand()%10;
        int b=rand()%10;
        if((a*a+b*b)<=100)
            count++;
        i--;
    }
    printf("%d",count);
}

8、数组a[N]，存放了1至N-1个数，其中某个数重复一次，找出重复的那个数

 

 

#include<iostream>  
using namespace std;  
void do_dup(int a[] , int n)  
{  
    int *b = new int[n];  
    for( int i = 0 ; i != n ; ++i )  
    {  
        b[i] = -1 ;  
    }  
    for( int j = 0 ; j != n ; ++j )  
    {  
        if( b[a[j]] == -1 ){  
            b[a[j]] = a[j] ;  
        }  
        else  
        {  
        cout << b[ a[j] ] << endl ;  
        break ;  
        }    
    }  
}  
int main(){  
    int a[]={  1 , 2 , 3 , 4 , 3 } ;  
    do_dup( a , 5 ) ;  
}   

9、用两个线程实现1-100的输出

 

package lzf.thread;
//用两个线程实现1-100的输出
public class Synchronized {
	// state==1表示线程1开始打印，state==2表示线程2开始打印
	private static int state = 1;

	private static int num1 = 1;
	private static int num2 = 1;

	public static void main(String[] args) {
		final Synchronized t = new Synchronized();
		new Thread(new Runnable() {
			@Override
			public void run() {
				while (num1 < 95){
					// 两个线程都用t对象作为锁，保证每个交替期间只有一个线程在打印
					synchronized (t) {
						// 如果state!=1, 说明此时尚未轮到线程1打印, 线程1将调用t的wait()方法, 直到下次被唤醒
						if (state != 1) {
							try {
								t.wait();
							} catch (InterruptedException e) {
								e.printStackTrace();
							}
						}
						// 当state=1时, 轮到线程1打印5次数字
						for (int j = 0; j < 5; j++) {
							System.out.println("num1:"+num1);
							num1 += 1;
							num2 = num1;
						}
						// 线程1打印完成后, 将state赋值为2, 表示接下来将轮到线程2打印
						state = 2;
						// notifyAll()方法唤醒在t上wait的线程2, 同时线程1将退出同步代码块, 释放t锁
						t.notifyAll();
					}
				}
			}
		}).start();

		new Thread(new Runnable() {
			@Override
			public void run() {
				while (num2 < 100) {
					synchronized (t) {
						if (state != 2) {
							try {
								t.wait();
							} catch (InterruptedException e) {
								e.printStackTrace();
							}
						}
						for (int j = 0; j < 5; j++) {
							System.out.println("num2:"+num2);
							num2 += 1;
							num1 = num2;
						}
						state = 1;
						t.notifyAll();
					}
				}
			}
		}).start();
	}
}

linux下线程示例

gcc -o pthread_testpthread_test .c -lpthread

 

#include<stddef.h>
#include<stdio.h>
#include<unistd.h>
#include"pthread.h"

void reader_function(void);
void writer_function(void);
char buffer;
int buffer_has_item=0;
pthread_mutex_t mutex;
main()
{
    pthread_t reader;
    pthread_mutex_init(&mutex,NULL);
    pthread_create(&reader,NULL,(void*)&reader_function,NULL);
    writer_function();
}
void writer_function(void)
{
    while(1)
    {
        pthread_mutex_lock(&mutex);
        if(buffer_has_item==0)
        {
	    			buffer='a';
            printf("make a new item\n");
            buffer_has_item=1;
        }
        pthread_mutex_unlock(&mutex);
    }
}
void reader_function(void)
{
    while(1)
    {
        pthread_mutex_lock(&mutex);
        if(buffer_has_item==1)
        {
	    			buffer='\0';
            printf("consume  item\n");
            buffer_has_item=0;
        }
        pthread_mutex_unlock(&mutex);
    }
}

 

线程模拟火车售票

 

package lzf.thread;

class TicketSystem {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		SellThread st = new SellThread();
		new Thread(st).start();
		try{
			Thread.sleep(10);
		}
		catch (Exception e) {
			// TODO: handle exception
			e.printStackTrace();
		}
		new Thread(st).start();
		st.b = true;
		
	}

}

class SellThread implements Runnable{
	int tickets = 100;
	Object obj = new Object();
	boolean b = false;
	@Override
	public void run() {
		// TODO Auto-generated method stub
		if(b==false){
			while(true){
				sell();
			}
		}
		while (true) {
			synchronized (this) {
				if(tickets>0){
					try{
						Thread.sleep(1);
					}
					catch (Exception e) {
						// TODO: handle exception
						e.printStackTrace();
					}
					System.out.println("obj"+Thread.currentThread().getName()+" sell tickets:"+tickets);
					tickets--;
				}
			}
		}
	}
	public synchronized void sell(){
		if(tickets>0){
			try{
				Thread.sleep(10);
			}
			catch (Exception e) {
				// TODO: handle exception
				e.printStackTrace();
			}
			System.out.println("sell"+Thread.currentThread().getName()+" sell tickets:"+tickets);
			tickets--;
		}
	}
}

 

10、宏定义交换两个数字

 

//第一种
#define SWAP(x,y) ((x)=(x)+(y),(y)=(x)-(y),(x)=(x)-(y))
//第二种
#define SWAP(x,y) ((x)=(x)^(y),(y)=(x)^(y),(x)=(x)^(y))//比上一种更好，不会出现大数字的溢出问题
#define swap(x, y)/
//带有换行
x = x + y;/
y = x - y;/
x = x - y;

#define swap(x, y)/
x ^= y;/
y ^= x;/
x ^= y;
void main()
{
int x=3,y=4;
swap(x,y);
printf("%d,%d",x,y);
}