读取文件中信息，包括动态字符串数组，C++premier plus的一道课后题

[心态与习惯]

该文件的第一项为捐款人数，余下的内容应为成对的行。在每一对中，第一行为捐款人姓名，第二行为捐款数额。即该文件类似于下面：
4
Sam Stone
2000
Freida Flass
100500
Tammy Tubbs
5000
Rich Raptor
55000
假设该文件为“123.txt”
读取该文件信息，并按捐款额是否大于15000分成两类输出

[荔枝树_12w]

用两个map容器，分别存储大于或小于的人，然后遍历容器输出。

[心态与习惯]

引用 3 楼 JiMoKuangXiangQu 的回复:

[quote=引用 楼主 robert_chen1988 的回复:] 该文件的第一项为捐款人数，余下的内容应为成对的行。在每一对中，第一行为捐款人姓名，第二行为捐款数额。即该文件类似于下面： 4 Sam Stone 2000 Freida Flass 100500 Tammy Tubbs 5000 Rich Raptor 55000 假设该文件为“123.txt” 读取该文件信息，并按捐款额是否大于15000分成两类输出

#include <iostream>
#include <string>
#include <vector>
#include <fstream>

using namespace std;


class DonorInfo {
	friend istream& operator>>(istream&, DonorInfo&);
	friend ostream& operator<<(ostream&, const DonorInfo&);

public:
	DonorInfo() : m_name(""), m_numOfDonations(0UL) { }

	unsigned long getDonation() const { return m_numOfDonations; }

private:
	string m_name;
	unsigned long m_numOfDonations;
};

istream& operator>>(istream& is, DonorInfo& donorInfo)
{
	string garbage;
	getline(is, garbage);

	getline(is, donorInfo.m_name);
	is >> donorInfo.m_numOfDonations;

	return is;
}

ostream& operator<<(ostream& os, const DonorInfo& donorInfo)
{
	os << donorInfo.m_name << ": " << donorInfo.m_numOfDonations;
	return os;
}


class DonorDB {
	friend ostream& operator<<(ostream&, const DonorDB&);

public:
	DonorDB(istream& is) {
		is >> m_numOfDonors;

		DonorInfo donorInfo;
		while (is >> donorInfo) {
			if (donorInfo.getDonation() <= 15000)
				m_donorsA.push_back(donorInfo);
			else
				m_donorsB.push_back(donorInfo);
		}
	}

private:
	unsigned int m_numOfDonors;
	vector<DonorInfo> m_donorsA, m_donorsB;
};

ostream& operator<<(ostream& os, const DonorDB& donorDB)
{
	os << donorDB.m_numOfDonors << endl;

	cout << endl << "Donors A( <= 15000): " << endl;
	for (vector<DonorInfo>::const_iterator it = donorDB.m_donorsA.begin(); 
		 it != donorDB.m_donorsA.end(); ++it)
		 os << *it << endl;

	cout << endl << "DonorsB( > 15000): " << endl;
	for (vector<DonorInfo>::const_iterator it = donorDB.m_donorsB.begin(); 
		 it != donorDB.m_donorsB.end(); ++it)
		 os << *it << endl;

	return os;
}



int main()
{
	ifstream ifile("123.txt");
	DonorDB donorDB(ifile);
	cout << donorDB;
	ifile.close();

	cin.get();
	return 0;
}

仅简单的完成逻辑，没有做错误检测等等，仅供参考，也未必满足LZ的限制条件。 Input(123.txt):

4
Sam Stone
2000
Freida Flass
100500
Tammy Tubbs
5000
Rich Raptor
55000

Ouput(console):

4

Donors A( <= 15000):
Sam Stone: 2000
Tammy Tubbs: 5000

DonorsB( > 15000):
Freida Flass: 100500
Rich Raptor: 55000

[/quote]赞一个，但这个章节还没接触到类呢

[心态与习惯]

这个章节还没学到map容器，需用别的手段了

[JiMoKuangXiangQu]

引用 楼主 robert_chen1988 的回复:

该文件的第一项为捐款人数，余下的内容应为成对的行。在每一对中，第一行为捐款人姓名，第二行为捐款数额。即该文件类似于下面： 4 Sam Stone 2000 Freida Flass 100500 Tammy Tubbs 5000 Rich Raptor 55000 假设该文件为“123.txt” 读取该文件信息，并按捐款额是否大于15000分成两类输出

#include <iostream>
#include <string>
#include <vector>
#include <fstream>

using namespace std;


class DonorInfo {
	friend istream& operator>>(istream&, DonorInfo&);
	friend ostream& operator<<(ostream&, const DonorInfo&);

public:
	DonorInfo() : m_name(""), m_numOfDonations(0UL) { }

	unsigned long getDonation() const { return m_numOfDonations; }

private:
	string m_name;
	unsigned long m_numOfDonations;
};

istream& operator>>(istream& is, DonorInfo& donorInfo)
{
	string garbage;
	getline(is, garbage);

	getline(is, donorInfo.m_name);
	is >> donorInfo.m_numOfDonations;

	return is;
}

ostream& operator<<(ostream& os, const DonorInfo& donorInfo)
{
	os << donorInfo.m_name << ": " << donorInfo.m_numOfDonations;
	return os;
}


class DonorDB {
	friend ostream& operator<<(ostream&, const DonorDB&);

public:
	DonorDB(istream& is) {
		is >> m_numOfDonors;

		DonorInfo donorInfo;
		while (is >> donorInfo) {
			if (donorInfo.getDonation() <= 15000)
				m_donorsA.push_back(donorInfo);
			else
				m_donorsB.push_back(donorInfo);
		}
	}

private:
	unsigned int m_numOfDonors;
	vector<DonorInfo> m_donorsA, m_donorsB;
};

ostream& operator<<(ostream& os, const DonorDB& donorDB)
{
	os << donorDB.m_numOfDonors << endl;

	cout << endl << "Donors A( <= 15000): " << endl;
	for (vector<DonorInfo>::const_iterator it = donorDB.m_donorsA.begin(); 
		 it != donorDB.m_donorsA.end(); ++it)
		 os << *it << endl;

	cout << endl << "DonorsB( > 15000): " << endl;
	for (vector<DonorInfo>::const_iterator it = donorDB.m_donorsB.begin(); 
		 it != donorDB.m_donorsB.end(); ++it)
		 os << *it << endl;

	return os;
}



int main()
{
	ifstream ifile("123.txt");
	DonorDB donorDB(ifile);
	cout << donorDB;
	ifile.close();

	cin.get();
	return 0;
}

仅简单的完成逻辑，没有做错误检测等等，仅供参考，也未必满足LZ的限制条件。 Input(123.txt):

4
Sam Stone
2000
Freida Flass
100500
Tammy Tubbs
5000
Rich Raptor
55000

Ouput(console):

4

Donors A( <= 15000):
Sam Stone: 2000
Tammy Tubbs: 5000

DonorsB( > 15000):
Freida Flass: 100500
Rich Raptor: 55000