程序设计实习——STL1

A:白给的list排序

总时间限制: 1000ms 内存限制: 65536kB

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

输入

输出

9.8,7.3,3.4,2.6,1.2,

样例输入

样例输出

同输入

来源

Guo Wei

B:按距离排序

总时间限制: 1000ms 内存限制: 65536kB

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 {
    T1 n;
    T2 x;
    Closer(T1 nn, T2 xx){
        //cout<<nn<<endl; 
        n=nn;x=xx;
    }
bool operator()(T1 &a, T1 &b){
    //cout<<a<<" "<<b<<endl;
    if(x(n,a)==x(n,b)){
        return a<b; 
    }
    return x(n,a)<x(n,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;
}

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

C:goodcopy

总时间限制: 1000ms 内存限制: 65536kB

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 {
void operator()(T* start, T* end, T* aim){
        int num=end-start;
        T* mid=new T[num];
        for(int i=0;start+i!=end;++i)
        mid[i]=start[i];
        for(int i=0;i<num;i++)
        aim[i]=mid[i];
        delete[] mid;
    }
};

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

D:函数对象的过滤器

总时间限制: 1000ms 内存限制: 65536kB

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;
    }
};
template <class T>
struct FilterClass{
    int m,n;
    FilterClass(int mm, int nn){
        m=mm;
        n=nn;
    }
    bool operator()(T &x){
        return T(m)<x&&x<T(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;
}

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

E:很难蒙混过关的CArray3d三维数组模板类

总时间限制: 1000ms 内存限制: 65536kB

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
{
public:
    class CArray2D{
    public:
        T* arr;
        int n,k;
        CArray2D(){
            n=k=0;
            arr=nullptr;
        }
        CArray2D(int jj,int kk):n(jj),k(kk){
            arr=new T[jj*kk];
        }
        T* operator[](int j){
            return arr+j*k;
        }
        operator T*(){
            return arr;
        }

    };
    CArray2D *arr;
    int m;
    int n;
    int k;
    CArray3D(int ii,int jj, int kk):m(ii),n(jj),k(kk) {
        arr = new CArray2D[ii];
        for(int i=0;i<ii;i++){
            arr[i].n=n;
            arr[i].k=k;
            arr[i].arr=new T[n*k];
        }
    }
    CArray2D& operator[](int i){
        return arr[i];
    }
};

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;
}

输入

输出

等同于样例

样例输入

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. ai[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

来源

Guo Wei

F:我自己的 ostream_iterator

总时间限制: 1000ms 内存限制: 65536kB

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
{
ostream & os_;
    string s_;
    T val;
public:
    myostream_iteraotr(ostream & os, const string &s) :os_(os), s_(s) {}

    T& operator *(){
        return val;
    }

    ostream&operator++(){
        os_<<val<<s_;
        return os_;
    }
};


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

输出

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

样例输入

样例输出

同输入

来源

Guo Wei

最后修改:2019 年 05 月 06 日 02 : 52 PM
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