18.04.04 作业STL 1

A:白给的list排序

描述

程序填空，产生指定输出

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

输入

无

输出

9.8,7.3,3.4,2.6,1.2,

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <list>
 5 using namespace std;
 6 int main()
 7 {    
 8     double a[] = {1.2,3.4,9.8,7.3,2.6};
 9     list<double> lst(a,a+5);
10     lst.sort(
11 [](double a, double b) -> bool { return a > b; }
12 );
13     
14     for(list<double>::iterator i  = lst.begin(); i != lst.end(); ++i) 
15         cout << * i << "," ;
16     return 0;
17 }

View Code

B:按距离排序

描述

程序填空，输出指定结果

#include <iostream>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
template <class T1,class T2>
struct Closer {
// 在此处补充你的代码
};

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;
}

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,

来源Guo Wei

 1 #include <iostream>
 2 #include <cmath>
 3 #include <algorithm>
 4 #include <string>
 5 using namespace std;
 6 template <class T1,class T2>
 7 struct Closer {
 8     T1 n;
 9     T2 d;
10     Closer(T1 a, T2 d1) :n(a), d(d1) {}
11     bool operator()(const T1&a, const T1&b) {
12         if (d(a, n) == d(b,n))
13             return a < b;
14         else return d(a,n)<d(b,n);
15     }
16 };
17 
18 int Distance1(int n1,int n2) {
19     return abs(n1-n2);
20 }
21 int Distance2(const string & s1, const string & s2)
22 {
23     return abs((int)s1.length()- (int) s2.length());
24 }
25 int a[10] = { 0,3,1,4,7,9,20,8,10,15};
26 string b[6] = {"American","Jack","To","Peking","abcdefghijklmnop","123456789"};
27 int main()
28 {
29     int n;string s;
30     while( cin >> n >> s ) {
31         sort(a,a+10,Closer<int ,int (*)(int ,int)> (n,Distance1));
32         for(int i = 0;i < 10; ++i)
33             cout << a[i] << "," ;
34         cout << endl;
35         sort(b,b+6,Closer<string,int (*)(const string &,const string &  )> (s,Distance2)); 
36         for(int i = 0;i < 6; ++i)
37             cout << b[i] << "," ;
38         cout << endl;
39     }
40     return 0;
41 }

View Code

C:goodcopy

描述

编写GoodCopy类模板，使得程序按指定方式输出

#include <iostream>
using namespace std;


template <class T>
struct GoodCopy {
// 在此处补充你的代码
};

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;
}

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,

来源Guo Wei

 1 #include <iostream>
 2 using namespace std;
 3 
 4 
 5 template <class T>
 6 struct GoodCopy {
 7     void cpy(T*s,T*e,T*to) {
 8         for (T*i = e-1; i >=s; i--)
 9             *(to + (i - s)) = *i;
10     }
11     void operator()(T*s,T*e,T*to) {
12         cpy(s, e, to);
13     }
14 };
15 
16 int a[200];
17 int b[200];
18 string c[200];
19 string d[200];
20 
21 template <class T>
22 void Print(T s,T e) {
23     for(; s != e; ++s)
24         cout << * s << ",";
25     cout << endl;
26 }
27 
28 int main()
29 {
30     int t;
31     cin >> t;
32     while( t -- ) {
33         int m ;
34         cin >> m;
35         for(int i = 0;i < m; ++i)
36             cin >> a[i];
37         GoodCopy<int>()(a,a+m,b);
38         Print(b,b+m);
39         GoodCopy<int>()(a,a+m,a+m/2);
40         Print(a+m/2,a+m/2 + m);
41 
42         for(int i = 0;i < m; ++i)
43             cin >> c[i];
44         GoodCopy<string>()(c,c+m,d);
45         Print(c,c+m);
46         GoodCopy<string>()(c,c+m,c+m/2);
47         Print(c+m/2,c+m/2 + m);
48     }
49     return 0;
50 }

View Code

D:函数对象的过滤器

描述

程序填空输出指定结果

#include <iostream>
#include <vector>
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>
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;
}

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,

来源Guo Wei

 1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 4 
 5 
 6 struct A {
 7     int v;
 8     A() { }
 9     A(int n):v(n) { };
10     bool operator<(const A & a) const {
11         return v < a.v;
12     }
13 };
14 template<class T1>
15 struct FilterClass {
16     T1 m, n;
17     FilterClass(T1 a, T1 b):m(a), n(b) {}
18     bool operator()(T1 s) {
19         return (m < s) && (s < n);
20     }
21 };
22 
23 template <class T>
24 void Print(T s,T e)
25 {
26     for(;s!=e; ++s)
27         cout << *s << ",";
28     cout << endl;
29 }
30 template <class T1, class T2,class T3>
31 T2 Filter( T1 s,T1 e, T2 s2, T3 op) 
32 {
33     for(;s != e; ++s) {
34         if( op(*s)) {
35             * s2 = * s;
36             ++s2;
37         }
38     }
39     return s2;
40 }
41 
42 ostream & operator <<(ostream & o,A & a)
43 {
44     o << a.v;
45     return o;
46 }
47 vector<int> ia;
48 vector<A> aa; 
49 int main()
50 {
51     int m,n;
52     while(cin >> m >> n) {
53         ia.clear();
54         aa.clear(); 
55         int k,tmp;
56         cin >> k;
57         for(int i = 0;i < k; ++i) {
58             cin >> tmp; 
59             ia.push_back(tmp);
60             aa.push_back(tmp); 
61         }
62         vector<int> ib(k);
63         vector<A> ab(k);
64         vector<int>::iterator p =  Filter(ia.begin(),ia.end(),ib.begin(),FilterClass<int>(m,n));
65         Print(ib.begin(),p);
66         vector<A>::iterator pp = Filter(aa.begin(),aa.end(),ab.begin(),FilterClass<A>(m,n));
67         Print(ab.begin(),pp);
68         
69     }
70     return 0;
71 }

View Code

E:很难蒙混过关的CArray3d三维数组模板类

描述

实现一个三维数组模版CArray3D，可以用来生成元素为任意类型变量的三维数组，输出指定结果

#include <iostream>
#include <iomanip> 
#include <cstring>
using namespace std;
template <class T>
class CArray3D
{
// 在此处补充你的代码
};

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;
}

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

 1 #include <iostream>
 2 #include <iomanip> 
 3 #include <cstring>
 4 using namespace std;
 5 template <class T>
 6 class CArray3D
 7 {
 8 public:
 9     class CArray2D {
10     public:
11         int y, z;
12         T *p;
13         void set(int m, int n) {
14             y = m, z = n;
15             p = new T[m*n];
16         }
17         T*operator[](int k) {
18             return p+k*z;
19         }
20         operator T*() {
21             return this->p;
22         }
23     };
24     int x;
25     CArray2D *ptoa;
26     CArray3D(int m, int n, int k) :x(m) {
27         ptoa = new CArray2D[x];
28         for (int i = 0; i < x; i++)
29             ptoa[i].set(n, k);
30     }
31     CArray2D&operator[](int k) {
32         return ptoa[k];
33     }
34 };
35 
36 CArray3D<int> a(3,4,5);
37 CArray3D<double> b(3,2,2);
38 void PrintA()
39 {
40     for(int i = 0;i < 3; ++i) {
41         cout << "layer " << i << ":" << endl;
42         for(int j = 0; j < 4; ++j) {
43             for(int k = 0; k < 5; ++k) 
44                 cout << a[i][j][k] << "," ;
45             cout << endl;
46         }
47     }
48 }
49 void PrintB()
50 {
51     for(int i = 0;i < 3; ++i) {
52         cout << "layer " << i << ":" << endl;
53         for(int j = 0; j < 2; ++j) {
54             for(int k = 0; k < 2; ++k) 
55                 cout << b[i][j][k] << "," ;
56             cout << endl;
57         }
58     }
59 }
60 
61 int main()
62 {
63 
64     int No = 0;
65     for( int i = 0; i < 3; ++ i ) {
66         a[i];
67         for( int j = 0; j < 4; ++j ) {
68             a[j][i];
69             for( int k = 0; k < 5; ++k )
70                 a[i][j][k] = No ++;
71             a[j][i][i];    
72         }
73     }
74     PrintA();
75     memset(a[1],-1 ,20*sizeof(int));    
76     memset(a[1],-1 ,20*sizeof(int));
77     PrintA(); 
78     memset(a[1][1],0 ,5*sizeof(int));    
79     PrintA();
80 
81     for( int i = 0; i < 3; ++ i )
82         for( int j = 0; j < 2; ++j )
83             for( int k = 0; k < 2; ++k )
84                 b[i][j][k] = 10.0/(i+j+k+1);
85     PrintB();
86     int n = a[0][1][2];
87     double f = b[0][1][1];
88     cout << "****" << endl;
89     cout << n << "," << f << endl;
90         
91     return 0;
92 }

View Code

内部类是这个意思的吗还有这样的

F:我自己的 ostream_iterator

描述

程序填空输出指定结果

#include <iostream>
#include <list>
#include <string>
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
};


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;
}

输入

无

输出

5,21,14,2,3,
1.4--5.56--3.2--98.3--3.3--

 1 #include <iostream>
 2 #include <list>
 3 #include <string>
 4 using namespace std;
 5 
 6 template <class T1,class T2>
 7 void Copy(T1 s,T1 e, T2 x)
 8 {
 9     for(; s != e; ++s,++x)
10         *x = *s;
11 }
12 
13  
14 template<class T>
15 class myostream_iteraotr
16 {
17 public:
18     string sep;
19     ostream&os;
20     void operator++() {}
21     myostream_iteraotr(ostream&out, string s) :os(out), sep(s) {}
22     myostream_iteraotr&operator*() {
23         return *this;
24     }
25     myostream_iteraotr&operator=(T&s) {
26         os << s<<sep;
27         return *this;
28     }
29 };
30 
31 
32 int main()
33 {    const int SIZE = 5;
34     int a[SIZE] = {5,21,14,2,3};
35     double b[SIZE] = { 1.4, 5.56,3.2,98.3,3.3};
36     list<int> lst(a,a+SIZE);
37     myostream_iteraotr<int> output(cout,",");
38     Copy( lst.begin(),lst.end(),output); 
39     cout << endl;
40     myostream_iteraotr<double> output2(cout,"--");
41     Copy(b,b+SIZE,output2);
42     return 0;
43 }

View Code

注定失败的战争，也要拼尽全力去打赢它；就算输，也要输得足够漂亮。