Question

我是c ++的初学者现在我被困在练习上，我不知道如何解决它

练习的目标是：写一个程序，要求用户输入由10个不同的人吃早餐的煎饼数量，并制作从最高到最小的列表，如：人6吃：10个煎饼人1吃：6煎饼等...

我有这段代码：

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{

    int person_num = 0;
    int pancakes;
    vector<int> pancake_list;


    while (true) {

        person_num = person_num + 1;
        cout << "Please enter the number of pancakes eaten by Person #" << person_num << " ";
        cin >> pancakes;
        pancake_list.push_back(pancakes);

        sort(begin(pancake_list), end(pancake_list));
        reverse(begin(pancake_list), end(pancake_list));

        if (person_num == 10) {
            system("cls");
            for (int i = 0; i < pancake_list.size(); i++) {
                cout << pancake_list[i] << endl;
            }
            system("pause");
        }
    }


    return 0;
}

问题在于我不知道如何将分类和反转的煎饼分配给合适的人请帮忙解释

抱歉我的英文

Answer 1

我会对c++ map做点什么。如果您知道要对事物进行排序，则可以使用地图键排序来使您的生活更轻松。在这种情况下，使用在地图中作为关键字吃的煎饼数量，以及人数作为值 - 列表已经排序的方式，您可以向后迭代以反向排序顺序打印它。像这样：

#include "stdafx.h"
#include <iostream>
#include <map>

using namespace std;

int main()
{

int person_num = 0;
int pancakes;
multimap<int, int> pancake_map;// Use multimap incase two people ate the same number of pancakes. Note: there are no guarantees about the relative order of elements with the same key.


while (true) {

        person_num = person_num + 1;
        cout << "Please enter the number of pancakes eaten by Person #" << person_num << " ";
        cin >> pancakes;
        pancake_map.insert(make_pair(pancakes,person_num));// c++ multimap stores things sorted by key value, so we don't have to do any sorting

        if (person_num == 10) 
        {
            system("cls");
            // to print the ascending sorted list, iterate through the container forwards
            for (multimap<int,int>::iterator it=pancake_map.begin(); it != pancake_map.end(); ++it) 
            {
                cout << "Person number " << it->second << " ate " << it->first << " pancakes." << endl;
            }

            cout << endl;// little bit of formatting...
            // to print the reverse sorted list, iterate backwards
            for (multimap<int,int>::reverse_iterator it=pancake_map.rbegin(); it != pancake_map.rend(); ++it) 
            {
                cout << "Person number " << it->second << " ate " << it->first << " pancakes." << endl;
            }


            system("pause");
            break; // I added this to escape the loop after entering 10 people, remove it if you want.
        }
    }


    return 0;
}

Answer 2

我认为可以使用数组而不是向量，因为元素的数量很少并且是固定的。

其中一个解决方案可以采用以下方式

#include <iostream>
#include <utility>
#include <algorithm>

int main() 
{
    const size_t N = 10;
    std::pair<size_t, size_t> pancake_list[N];

    std::cout << "Please enter the number of pancakes eaten by "
              << N << " persons: ";

    size_t n = 0;

    for ( ; n < N && std::cin >> pancake_list[n].second; n++ )
    {
        pancake_list[n].first = n + 1;
    }

    std::sort( pancake_list, pancake_list + n,
               []( auto &a, auto &b ) { return b.second < a.second; } );


    for ( size_t i = 0; i < n; i++ )
    {
        std::cout << "Person #" << pancake_list[i].first 
                  << " ate " << pancake_list[i].second
                  << " pancakes." << std::endl;
    }


    return 0;
}

例如，如果要输入以下煎饼序列

6 2 8 7 1 4 6 3 5 0

然后输出看起来像

Person #4 ate 7 pancakes.
Person #1 ate 6 pancakes.
Person #7 ate 6 pancakes.
Person #9 ate 5 pancakes.
Person #6 ate 4 pancakes.
Person #8 ate 3 pancakes.
Person #2 ate 2 pancakes.
Person #5 ate 1 pancakes.
Person #10 ate 0 pancakes.

Answer 3

到目前为止，其他解决方案是使用一对（人＃，煎饼）整数。这是一个解决方案。另一个是要意识到你的问题是由于你没有遵守原始订单。对结果进行排序时，您使用新订单替换了原始订单。但如果您对副本进行排序，您仍然可以使用原始订单，并且可以匹配这两个。

以您的代码为基础：

// Ask the use how many pancakes
vector<int> pancake_copy = pancake_list;

sort(begin(pancake_copy), end(pancake_copy));
reverse(begin(pancake_copy), end(pancake_copy));

// You know have the sorted and the unsorted lists. 
for (int numPancakes : pancake_copy)
{
   std::cout << numPancakes " pancakes were eaten by: ";
   for (int person = 0; person != pancake_list.size(); ++person)
   {
      if (pancakes_list[person] == numPancakes)
      {
         // This person ate numPancakes
         std::cout << i << " ";
      }
   }
   std::cout << std::endl;
}

奖金锻炼：当两个人吃相同数量的煎饼时，现在出了什么问题？您可以使用哪种C ++函数来解决此问题？

C ++练习煎饼

3 个答案: