程序设计与算法（三）第08周测验(2020春季)

001:goodcopy

输入
第一行是整数 t,表示数据组数
每组数据：
第一行是整数 n , n < 50
第二行是 n 个整数
第三行是 n 个字符串
输出
将输入的整数原序输出两次，用","分隔
然后将输入的字符串原序输出两次，也用 ","分隔
样例输入
2
4
1 2 3 4
Tom Jack Marry Peking
1
0
Ted
样例输出
1,2,3,4,
1,2,3,4,
Tom,Jack,Marry,Peking,
Tom,Jack,Marry,Peking,
0,
0,
Ted,
Ted,

#include 
using namespace std;


template <class T>
struct GoodCopy {
	// 在此处补充你的代码
	void operator()(T* s, T* e, T* b) {
		int len = e - s; T tmp[200];
		for (int i = 0; i < len; i++) tmp[i] = s[i];
		for (int i = 0; i < len; i++) b[i] = tmp[i];
	}
//——————————————————————————————————————————
};

int a[200];
int b[200];
string c[200];
string d[200];

template <class T>
void Print(T s, T e) {
	for (; s != e; ++s)
		cout << *s << ",";
	cout << endl;
}

int main()
{
	int t;
	cin >> t;
	while (t--) {
		int m;
		cin >> m;
		for (int i = 0; i < m; ++i)
			cin >> a[i];
		GoodCopy<int>()(a, a + m, b);
		Print(b, b + m);
		GoodCopy<int>()(a, a + m, a + m / 2);
		Print(a + m / 2, a + m / 2 + m);

		for (int i = 0; i < m; ++i)
			cin >> c[i];
		GoodCopy<string>()(c, c + m, d);
		Print(c, c + m);
		GoodCopy<string>()(c, c + m, c + m / 2);
		Print(c + m / 2, c + m / 2 + m);
	}
	return 0;
}

002:按距离排序

输入
多组数据，每组一行，是一个整数n和一个字符串s
输出
定义两个整数的距离为两个整数差的绝对值
定义两个字符串的距离为两个字符串长度差的绝对值
对每组数据：
对数组a按和n的距离从小到大排序后输出。距离相同的，值小的排在前面。
然后对数组b，按照和s的距离从小到大输出。距离相同的，字典序小的排在前面
样例输入
2 a123456
4 a12345
样例输出
1,3,0,4,7,8,9,10,15,20,
American,Peking,123456789,Jack,To,abcdefghijklmnop,
4,3,1,7,0,8,9,10,15,20,
Peking,American,Jack,123456789,To,abcdefghijklmnop,

#include 
#include 
#include 
#include 
using namespace std;
template <class T1, class T2>
struct Closer {
	// 在此处补充你的代码
	T1 n; T2 dis;
	Closer(T1 nn, T2 dd) :n(nn), dis(dd) {}
	bool operator ()(const T1& a, const T1& b) {
		int d1 = dis(a, n);
		int d2 = dis(b, n);
		if (d1 != d2)return d1 < d2;
		return a < b;
	}
//————————————————————————————————
};

int Distance1(int n1, int n2) {
	return abs(n1 - n2);
}
int Distance2(const string& s1, const string& s2)
{
	return abs((int)s1.length() - (int)s2.length());
}
int a[10] = { 0,3,1,4,7,9,20,8,10,15 };
string b[6] = { "American","Jack","To","Peking","abcdefghijklmnop","123456789" };
int main()
{
	int n; string s;
	while (cin >> n >> s) {
		sort(a, a + 10, Closer<int, int (*)(int, int)>(n, Distance1));
		for (int i = 0; i < 10; ++i)
			cout << a[i] << ",";
		cout << endl;
		sort(b, b + 6, Closer<string, int (*)(const string&, const string&)>(s, Distance2));
		for (int i = 0; i < 6; ++i)
			cout << b[i] << ",";
		cout << endl;
	}
	return 0;
}

003:很难蒙混过关的CArray3d三维数组模板类

输入
无
输出
等同于样例
样例输入
无
样例输出
layer 0:
0,1,2,3,4,
5,6,7,8,9,
10,11,12,13,14,
15,16,17,18,19,
layer 1:
20,21,22,23,24,
25,26,27,28,29,
30,31,32,33,34,
35,36,37,38,39,
layer 2:
40,41,42,43,44,
45,46,47,48,49,
50,51,52,53,54,
55,56,57,58,59,
layer 0:
0,1,2,3,4,
5,6,7,8,9,
10,11,12,13,14,
15,16,17,18,19,
layer 1:
-1,-1,-1,-1,-1,
-1,-1,-1,-1,-1,
-1,-1,-1,-1,-1,
-1,-1,-1,-1,-1,
layer 2:
40,41,42,43,44,
45,46,47,48,49,
50,51,52,53,54,
55,56,57,58,59,
layer 0:
0,1,2,3,4,
5,6,7,8,9,
10,11,12,13,14,
15,16,17,18,19,
layer 1:
-1,-1,-1,-1,-1,
0,0,0,0,0,
-1,-1,-1,-1,-1,
-1,-1,-1,-1,-1,
layer 2:
40,41,42,43,44,
45,46,47,48,49,
50,51,52,53,54,
55,56,57,58,59,
layer 0:
10,5,
5,3.33333,
layer 1:
5,3.33333,
3.33333,2.5,
layer 2:
3.33333,2.5,
2.5,2,
****
7,3.33333

提示
建议做法：

1. a[i][j][k] 这个表达式的第一个[]返回一个内部类的对象，该内部类也重载了[],且返回值为指针。
2. 必要时需重载对象到指针的强制类型转换运算符

#include 
#include  
#include 
using namespace std;
template <class T>
class CArray3D
{
	// 在此处补充你的代码
public:
	int i, j, k;
	T* p;
	CArray3D(int a, int b, int c) :i(a), j(b), k(c) {
		p = new T[i * j * k];
	}
	~CArray3D() { delete[] p; }
	class Ret {
	public:
		T* ip;
		int k;
		Ret(T* a, int b) :ip(a), k(b) {}
		T* operator [](int j) {//第二个[]
			return ip + j * k;
		}
		operator T* () { return ip; }//下面的代码中有a[1],a[1]的类型为Ret,函数形参为*void,因此要重载T*
	};
	Ret operator[](int i) {//第一个[]
		T* ip = p + i * j * k;
		return Ret(ip, k);
	}
};
CArray3D<int> a(3, 4, 5);
CArray3D<double> b(3, 2, 2);
void PrintA()
{
	for (int i = 0; i < 3; ++i) {
		cout << "layer " << i << ":" << endl;
		for (int j = 0; j < 4; ++j) {
			for (int k = 0; k < 5; ++k)
				cout << a[i][j][k] << ",";
			cout << endl;
		}
	}
}
void PrintB()
{
	for (int i = 0; i < 3; ++i) {
		cout << "layer " << i << ":" << endl;
		for (int j = 0; j < 2; ++j) {
			for (int k = 0; k < 2; ++k)
				cout << b[i][j][k] << ",";
			cout << endl;
		}
	}
}

int main()
{

	int No = 0;
	for (int i = 0; i < 3; ++i) {
		a[i];
		for (int j = 0; j < 4; ++j) {
			a[j][i];
			for (int k = 0; k < 5; ++k)
				a[i][j][k] = No++;
			a[j][i][i];
		}
	}
	PrintA();
	memset(a[1], -1, 20 * sizeof(int));
	memset(a[1], -1, 20 * sizeof(int));
	PrintA();
	memset(a[1][1], 0, 5 * sizeof(int));
	PrintA();

	for (int i = 0; i < 3; ++i)
		for (int j = 0; j < 2; ++j)
			for (int k = 0; k < 2; ++k)
				b[i][j][k] = 10.0 / (i + j + k + 1);
	PrintB();
	int n = a[0][1][2];
	double f = b[0][1][1];
	cout << "****" << endl;
	cout << n << "," << f << endl;

	return 0;
}

004:函数对象的过滤器

输入
多组数据
每组数据两行

第一行是两个整数 m 和 n
第二行先是一个整数k ,然后后面跟着k个整数
输出
对每组数据，按原顺序输出第二行的后k个整数中，大于m且小于n的数
输出两遍
数据保证一定能找到符合要求的整数
样例输入
1 3
1 2
2 8
5 1 2 3 4 9
样例输出
2,
2,
3,4,
3,4,

#include 
#include 
using namespace std;


struct A {
	int v;
	A() { }
	A(int n) :v(n) { };
	bool operator<(const A& a) const {
		return v < a.v;
	}
};
// 在此处补充你的代码
template<class T>
class FilterClass {
	int m, n;
public:
	FilterClass(int _m, int _n) :m(_m), n(_n) {}
	bool operator()(const T& a) {
		return T(m) < a&& a < n;
	}
};
//————————————————————————————————
template <class T>
void Print(T s, T e)
{
	for (; s != e; ++s)
		cout << *s << ",";
	cout << endl;
}
template <class T1, class T2, class T3>
T2 Filter(T1 s, T1 e, T2 s2, T3 op)
{
	for (; s != e; ++s) {
		if (op(*s)) {
			*s2 = *s;
			++s2;
		}
	}
	return s2;
}

ostream& operator <<(ostream& o, A& a)
{
	o << a.v;
	return o;
}
vector<int> ia;
vector<A> aa;
int main()
{
	int m, n;
	while (cin >> m >> n) {
		ia.clear();
		aa.clear();
		int k, tmp;
		cin >> k;
		for (int i = 0; i < k; ++i) {
			cin >> tmp;
			ia.push_back(tmp);
			aa.push_back(tmp);
		}
		vector<int> ib(k);
		vector<A> ab(k);
		vector<int>::iterator p = Filter(ia.begin(), ia.end(), ib.begin(), FilterClass<int>(m, n));
		Print(ib.begin(), p);
		vector<A>::iterator pp = Filter(aa.begin(), aa.end(), ab.begin(), FilterClass<A>(m, n));
		Print(ab.begin(), pp);

	}
	return 0;
}

005:白给的list排序

输入
无
输出
9.8,7.3,3.4,2.6,1.2,

#include 
#include 
#include 
#include 
using namespace std;
int main()
{
	double a[] = { 1.2,3.4,9.8,7.3,2.6 };
	list<double> lst(a, a + 5);
	lst.sort(
		// 在此处补充你的代码
		greater<double>()
//——————————————————————————
	);

	for (list<double>::iterator i = lst.begin(); i != lst.end(); ++i)
		cout << *i << ",";
	return 0;
}

006:我自己的 ostream_iterator

输入
无
输出
5,21,14,2,3,
1.4–5.56–3.2–98.3–3.3–

#include 
#include 
#include 
using namespace std;

template <class T1, class T2>
void Copy(T1 s, T1 e, T2 x)
{
	for (; s != e; ++s, ++x)
		*x = *s;
}


template<class T>
class myostream_iteraotr
{
	// 在此处补充你的代码
	string Oo;
public:
	myostream_iteraotr(ostream& os, string out) :Oo(out) {}
	myostream_iteraotr& operator *() { return *this; }
	void operator =(T s) { cout << s << Oo; }
	void operator ++() {}
//——————————————————————————————————————————
};


int main()
{
	const int SIZE = 5;
	int a[SIZE] = { 5,21,14,2,3 };
	double b[SIZE] = { 1.4, 5.56,3.2,98.3,3.3 };
	list<int> lst(a, a + SIZE);
	myostream_iteraotr<int> output(cout, ",");
	Copy(lst.begin(), lst.end(), output);
	cout << endl;
	myostream_iteraotr<double> output2(cout, "--");
	Copy(b, b + SIZE, output2);
	return 0;
}

007:List

描述
写一个程序完成以下命令：
new id ——新建一个指定编号为id的序列(id<10000)
add id num——向编号为id的序列加入整数num
merge id1 id2——合并序列id1和id2中的数，并将id2清空
unique id——去掉序列id中重复的元素
out id ——从小到大输出编号为id的序列中的元素，以空格隔开
输入
第一行一个数n，表示有多少个命令( n＜＝２０００００)。以后n行每行一个命令。
输出
按题目要求输出。
样例输入
16
new 1
new 2
add 1 1
add 1 2
add 1 3
add 2 1
add 2 2
add 2 3
add 2 4
out 1
out 2
merge 1 2
out 1
out 2
unique 1
out 1
样例输出
1 2 3
1 2 3 4
1 1 2 2 3 3 4
 
1 2 3 4

#include
#include
#include
using namespace std;
int main(void) {
	int n, pos, num;
	list<int>id[10000];
	cin >> n;
	string cmd;
	while (n--) {
		cin >> cmd >> pos;
		if (cmd == "new") {}
		else if (cmd == "add") {
			cin >> num; id[pos].push_back(num);
		}
		else if (cmd == "merge") {
			cin >> num; id[pos].merge(id[num]);
		}
		else if (cmd == "unique") {
			id[pos].sort(); id[pos].unique();
		}
		else {
			id[pos].sort();
			for (list<int>::iterator it = id[pos].begin(); it != id[pos].end(); ++it)
				cout << *it << " ";
			cout << endl;
		}
	}
	return 0;
}