Question

我有这个代码将十进制数转换为二进制数：

#include <iostream>
#include <windows.h>
#include <vector>
#include <algorithm>
using namespace std;

void space(){ cout << endl << endl << endl; }

int main(int argc, char const *argv[]){
    system("color 1F");
    unsigned long int n, copy_n;
    vector<int> v;

    cout << "# Decimal:   "; cin >> n; cout << endl; copy_n = n;

    while (n != 0){
        if (n % 2 == 0){ v.push_back(0); }else{ v.push_back(1); }
        n = int(n / 2);}

    cout << endl << "# Binary:   ";

    reverse(v.begin(), v.end());

    for (size_t i = 0; i < v.size(); i++){cout << v.at(i);}


    space(); system("Pause"); return 0;
}

......而这......

#include <iostream>
#include <windows.h>
using namespace std;

void space(){ cout << endl << endl << endl; }

int main(int argc, char const *argv[]){
    system("color 1F");
    unsigned long int n, copy_n, nr = 0 ;
    cout << "# Decimal:  "; cin >> n; copy_n = n; cout << endl;

    while (copy_n != 0){ nr++; copy_n = int(copy_n / 2); }

    int* v = new int[nr];

    for (int i = 0; i < nr; i++){if (n % 2 == 0){ v[i] = 0; }else{ v[i] = 1; }n = int(n / 2);}

    cout << endl << "# Binary:  ";

    for (int i = nr-1; i >= 0;i--){cout << v[i] << "";}

    space(); system("Pause"); return 0;}

......而这......

#include <iostream>
#include <windows.h>
#include <bitset>
using namespace std;

void space(){ cout << endl << endl << endl; }

int main(int argc, char const *argv[]){
    system("color 1F");
    unsigned int n;
    cout << "# Decimal:  "; cin >> n; cout << endl;
    bitset<16>binary(n);
    cout << endl << "# Binary:  " << binary << endl;

    space(); system("Pause"); return 0;
}

但我的问题是：

如何使用算法中的reverse（）函数并使用基于范围的for循环

打印矢量

例如：十进制= 2

带

reverse(v.begin(), v.end());
for (size_t i = 0; i < v.size(); i++){cout << v.at(i);}

程序打印10

带

reverse(v.begin(), v.end());
for(auto i : v){cout<<v.at(i);}

程序pritns 01

为什么呢？我该如何解决这个问题？

Answer 1

这句话

reverse(v.begin(), v.end());
for(auto i : v){cout<<v.at(i);}

完全错了。

有效代码看起来像

reverse(v.begin(), v.end());
for(auto i : v){ cout << i;}

此外，如果您为源代码中输入的符号数付款，则此声明

if (n % 2 == 0){ v.push_back(0); }else{ v.push_back(1); }

看起来非常好，因为它包含许多符号。否则最好写

v.push_back( n % 2 );

同样在你的一个程序中，你动态分配一个数组

int* v = new int[nr];

但不是免费的。在这种情况下，最好使用智能指针std::unique_ptr。

您也可以尝试编写递归函数。例如

#include <iostream>
#include <vector>

std::vector<unsigned int> & dec_to_bin( std::vector<unsigned int> &v, 
                                        unsigned int x )
{
    const unsigned int Base = 2;
    static size_t n;

    unsigned int digit = x % Base;
    ++n;

    if ( x /= Base )
    {
        dec_to_bin( v, x );
    }
    else
    {
        v.reserve( n );
        n = 0;
    }

    v.push_back( digit );

    return v;
}

int main() 
{
    while ( true )
    {
        std::cout << "Enter a non-negative number (0-exit): ";

        unsigned int x = 0;
        std::cin >> x;

        if ( !x ) break;

        std::vector<unsigned int> v;

        dec_to_bin( v, x );

        for ( auto digit : v ) std::cout << digit;
        std::cout << std::endl;
    }

    return 0;
}

如果要按顺序输入

然后程序输出

Enter a non-negative number (0-exit): 15
1111
Enter a non-negative number (0-exit): 7
111
Enter a non-negative number (0-exit): 3
11
Enter a non-negative number (0-exit): 1
1
Enter a non-negative number (0-exit): 0

使用基于范围的for循环以相反的顺序打印矢量

1 个答案: