描述

 

定义一个包含图书信息(书号、书名、价格)的顺序表,读入相应的图书数据来完成图书信息表的创建,然后将读入的图书信息逆序存储,逐行输出逆序存储后每本图书的信息。

输入

输入n+1行,第一行是图书数目n,后n行是n本图书的信息(书号、书名、价格),每本图书信息占一行,书号、书名、价格用空格分隔,价格之后没有空格。其中书号和书名为字符串类型,价格为浮点数类型。

输出

总计n行,第i行是原有图书表中第n-i+1行的图书的信息(书号、书名、价格),每本图书信息占一行,书号、书名、价格用空格分隔。其中价格输出保留两位小数。

输入样例 1 

8
9787302257646 Data-Structure 35.00
9787302164340 Operating-System 50.00
9787302219972 Software-Engineer 32.00
9787302203513 Database-Principles 36.00
9787810827430 Discrete-Mathematics 36.00
9787302257800 Data-Structure 62.00
9787811234923 Compiler-Principles 62.00
9787822234110 The-C-Programming-Language 38.00

输出样例 1

9787822234110 The-C-Programming-Language 38.00
9787811234923 Compiler-Principles 62.00
9787302257800 Data-Structure 62.00
9787810827430 Discrete-Mathematics 36.00
9787302203513 Database-Principles 36.00
9787302219972 Software-Engineer 32.00
9787302164340 Operating-System 50.00
9787302257646 Data-Structure 35.00
#include <iostream>
#include<iomanip>
using namespace std;
#define MAXSIZE 10000

typedef struct 
{
	string no;
	string name;
	double price;
}Book;

typedef struct
{
	Book *elem;
	int num;
}SqList;

void Aver(SqList &L)
{
	double aver=0,sum=0;int i=0;
	for(i=0;i<L.num;i++)
		sum=sum+L.elem[i].price;
	aver=sum/L.num;
	for(i=0;i<L.num;i++)
	{
		if(L.elem[i].price<aver)
			L.elem[i].price=1.2*L.elem[i].price;
		else if(L.elem[i].price>=aver)
			L.elem[i].price=1.1*L.elem[i].price;
	}
	cout<<setiosflags(ios::fixed)<<setprecision(2)<<aver<<endl;
}

void Sort(SqList &L)
{
	int i=0,j=0;
	for(i=0;i<L.num-1;i++)
		for(j=0;j<L.num-1-i;j++)
			if(L.elem[j].price<L.elem[j+1].price)
			{
				Book temp;
				temp=L.elem[j];
				L.elem[j]=L.elem[j+1];
				L.elem[j+1]=temp;
			}
}

void Input(SqList &L)
{
	int n;int i=0;
	cin>>n;
	string a,b;
	double c;
	while(i<n)
	{	
		cin>>a>>b>>c;
		if(i==n) break;
		L.elem[L.num].no=a;
		L.elem[L.num].name=b;
		L.elem[L.num].price=c;
		L.num++;
		i++;
	}
	return ;
}

void Output(SqList &L)
{
	int i;
	for(i=0;i<L.num-1;i++)
		cout<<L.elem[i].no<<" "<<L.elem[i].name<<" "<<setiosflags(ios::fixed)<<setprecision(2)<<L.elem[i].price<<endl;
	cout<<L.elem[L.num-1].no<<" "<<L.elem[L.num-1].name<<" "<<setiosflags(ios::fixed)<<setprecision(2)<<L.elem[L.num-1].price<<endl;
	return ;
}

void InitList(SqList &L)
{
	L.elem=new Book[MAXSIZE];
	if(!L.elem) return;
	L.num=0;
	return;
}

void Upsidedown(SqList &L)//数组逆序输出 
{
	int i;
	for(i=0;i<L.num/2;i++)//交换一半即可,到n/2 
	{//注意数组下标从0开始,包含n个元素的数组,最后一个元素的下标是n-1
		Book temp;
		temp=L.elem[i];
		L.elem[i]=L.elem[L.num-1-i];//数组第i个与第n-1-i个交换 //数学归纳法 
		L.elem[L.num-1-i]=temp; 
	}
		
 } 

int main()
{
	SqList L;
	InitList(L);
	Input(L); 
	Upsidedown(L);
	Output(L);
	return 0;
}

 

posted on 2020-02-15 02:02  海月CSDN  阅读(606)  评论(0编辑  收藏  举报