我正在做一个像这样的练习题：
编写一个程序，要求用户输入吃过的汉堡包的数量 由十个不同的人（p1，p2，......等）
一旦输入数据，程序必须分析数据并输出哪个人吃最少的汉堡包，并输出吃汉堡包最多的人。
我知道我的代码并不理想（至少可以这么说，但我决定完成它，所以我可以为这个问题找些答案。如果你愿意帮忙，不要害怕向我抛出一些更复杂的东西，这就是我学习的方式，对吧？
这是我的代码：
#include <iostream>
#include <string>
using namespace std;
int main()
{
int p1 = 0;
int p2 = 0;
int p3 = 0;
int p4 = 0;
int p5 = 0;
int p6 = 0;
int p7 = 0;
int p8 = 0;
int p9 = 0;
int p10 = 0;
cout << "How many pancakes did p1 eat?" << endl;
cin >> p1;
cout << "How many pancakes did p2 eat?" << endl;
cin >> p2;
cout << "How many pancakes did p3 eat?" << endl;
cin >> p3;
cout << "How many pancakes did p4 eat?" << endl;
cin >> p4;
cout << "How many pancakes did p5 eat?" << endl;
cin >> p5;
cout << "How many pancakes did p6 eat?" << endl;
cin >> p6;
cout << "How many pancakes did p7 eat?" << endl;
cin >> p7;
cout << "How many pancakes did p8 eat?" << endl;
cin >> p8;
cout << "How many pancakes did p9 eat?" << endl;
cin >> p9;
cout << "How many pancakes did p10 eat?" << endl;
cin >> p10;
// LARGE section of if statements incoming
// Test to see which person ate the least pancakes
if (p1 < p2 && p1 < p3 && p1 < p4 && p1 < p5 && p1 < p6 && p1 < p7 && p1 < p8 && p1 < p9 && p1 < p10)
{
cout << "p1 ate the least pancakes." << endl;
}
if (p2 < p1 && p2 < p3 && p2 < p4 && p2 < p5 && p2 < p6 && p2 < p7 && p2 < p8 && p2 < p9 && p2 < p10)
{
cout << "p2 ate the least pancakes." << endl;
}
if (p3 < p1 && p3 < p2 && p3 < p4 && p3 < p5 && p3 < p6 && p3 < p7 && p3 < p8 && p3 < p9 && p3 < p10)
{
cout << "p3 ate the least pancakes." << endl;
}
if (p4 < p1 && p4 < p2 && p4 < p3 && p4 < p5 && p4 < p6 && p4 < p7 && p4 < p8 && p4 < p9 && p4 < p10)
{
cout << "p4 ate the least pancakes." << endl;
}
if (p5 < p1 && p5 < p2 && p5 < p3 && p5 < p4 && p5 < p6 && p5 < p7 && p5 < p8 && p5 < p9 && p5 < p10)
{
cout << "p5 ate the least pancakes." << endl;
}
if (p6 < p1 && p6 < p2 && p6 < p3 && p6 < p4 && p6 < p5 && p6 < p7 && p6 < p8 && p6 < p9 && p6 < p10)
{
cout << "p6 ate the least pancakes." << endl;
}
if (p7 < p1 && p7 < p2 && p7 < p3 && p7 < p4 && p7 < p5 && p7 < p6 && p7 < p7 && p7 < p9 && p7 < p10)
{
cout << "p7 ate the least pancakes." << endl;
}
if (p8 < p1 && p8 < p2 && p8 < p3 && p8 < p4 && p8 < p5 && p8 < p6 && p8 < p7 && p8 < p9 && p8 < p10)
{
cout << "p8 ate the least pancakes." << endl;
}
if (p9 < p1 && p9 < p2 && p9 < p3 && p9 < p4 && p9 < p5 && p9 < p6 && p9 < p7 && p9 < p8 && p9 < p10)
{
cout << "p9 ate the least pancakes." << endl;
}
if (p10 < p1 && p10 < p2 && p10 < p3 && p10 < p4 && p10 < p5 && p10 < p6 && p10 < p7 && p10 < p8 && p10 < p9)
{
cout << "p10 ate the least pancakes." << endl;
}
// Test to see who ate the most pancakes
if (p1 > p2 && p1 > p3 && p1 > p4 && p1 > p5 && p1 > p6 && p1 > p7 && p1 > p8 && p1 > p9 && p1 > p10)
{
cout << "p1 ate the most pancakes." << endl;
}
if (p2 > p1 && p2 > p3 && p2 > p4 && p2 > p5 && p2 > p6 && p2 > p7 && p2 > p8 && p2 > p9 && p2 > p10)
{
cout << "p2 ate the most pancakes." << endl;
}
if (p3 > p1 && p3 > p2 && p3 > p4 && p3 > p5 && p3 > p6 && p3 > p7 && p3 > p8 && p3 > p9 && p3 > p10)
{
cout << "p3 ate the most pancakes." << endl;
}
if (p4 > p1 && p4 > p2 && p4 > p3 && p4 > p5 && p4 > p6 && p4 > p7 && p4 > p8 && p4 > p9 && p4 > p10)
{
cout << "p4 ate the most pancakes." << endl;
}
if (p5 > p1 && p5 > p2 && p5 > p3 && p5 > p4 && p5 > p6 && p5 > p7 && p5 > p8 && p5 > p9 && p5 > p10)
{
cout << "p5 ate the most pancakes." << endl;
}
if (p6 > p1 && p6 > p2 && p6 > p3 && p6 > p4 && p6 > p5 && p6 > p7 && p6 > p8 && p6 > p9 && p6 > p10)
{
cout << "p6 ate the most pancakes." << endl;
}
if (p7 > p1 && p7 > p2 && p7 > p3 && p7 > p4 && p7 > p5 && p7 > p6 && p7 > p7 && p7 > p9 && p7 > p10)
{
cout << "p7 ate the most pancakes." << endl;
}
if (p8 > p1 && p8 > p2 && p8 > p3 && p8 > p4 && p8 > p5 && p8 > p6 && p8 > p7 && p8 > p9 && p8 > p10)
{
cout << "p8 ate the most pancakes." << endl;
}
if (p9 > p1 && p9 > p2 && p9 > p3 && p9 > p4 && p9 > p5 && p9 > p6 && p9 > p7 && p9 > p8 && p9 > p10)
{
cout << "p9 ate the most pancakes." << endl;
}
if (p10 > p1 && p10 > p2 && p10 > p3 && p10 > p4 && p10 > p5 && p10 > p6 && p10 > p7 && p10 > p8 && p10 > p9)
{
cout << "p10 ate the most pancakes." << endl;
}
return 0;
}
答案 0 :(得分：2)
试试这个：
typedef std::vector<unsigned int> Pancake_Container;
const unsigned int MAXIMUM_PARTICIPANTS = 10;
Pancake Container pancake_quantities(MAXIMUM_PARTICIPANTS);
for (unsigned int i = 0U; i < MAXIMUM_PARTICIPANTS; ++i)
{
static const char prompt_text1[] = "\nHow many pancakes did participant #";
static const char prompt_ending[] = " eat? ";
cout.write(prompt_text1, sizeof(prompt_text1) - 1U);
cout << i;
cout.write(prompt_ending, sizeof(prompt_ending) - 1U));
unsigned int quantity = 0U;
cin >> quantity;
pancake_quantities[i] = quantity;
}
unsigned int max_person_index = 0U;
unsigned int max_pancakes_eaten = 0U;
unsigned int min_person_index = 0U;
unsigned int min_pancakes_eaten = MAX_UINT;
for (unsigned int i = 0U; i < MAXIMUM_PARTICIPANTS; ++i)
{
const unsigned int quantity = pancake_quantities[i];
if (quantity > max_pancakes_eaten)
{
max_person_index = i;
max_pancakes_eaten = quantity;
}
if (quantity < min_pancakes_eaten)
{
min_person_index = i;
min_pancakes_eaten = quantity;
}
}
cout << "Person #" << max_person_index << " ate " << max_pancakes_eaten << "\n";
cout << "Person #" << min_person_index << " ate " << min_pancakes_eaten << "\n";
这个允许排序来计算最大值和最小值。
struct Pancake_Info
{
unsigned int id;
unsigned int quantity;
Pancake_Info() : id(0), quantity(0)
{}
bool operator < (const Pancake_Info& other)
{
return quantity < other.quantity;
}
};
typedef std::vector<Pancake_Info> Pancake_Container;
Pancake_Container pancake_quantities;
for (unsigned int i = 0U; i < MAXIMUM_PARTICIPANTS; ++i)
{
static const char prompt_text1[] = "\nHow many pancakes did participant #";
static const char prompt_ending[] = " eat? ";
cout.write(prompt_text1, sizeof(prompt_text1) - 1U);
cout << i;
cout.write(prompt_ending, sizeof(prompt_ending) - 1U);
unsigned int quantity = 0U;
Pancake_Info p_i;
cin >> p_i.quantity;
p_i.id = i;
pancake_quantities.push_back(p_i);
}
std::sort(pancake_quantities.begin(), pancake_quantities.end());
cout << "Minimum of "
<< pancake_quantities[0].quantity
<< " eaten by person "
<< pancake_quantities[0].id
<< "\n";
cout << "Maximum of "
<< pancake_quantities[MAXIMUM_PARTICIPANTS - 1].quantity
<< " eaten by person "
<< pancake_quantities[MAXIMUM_PARTICIPANTS - 1].id
<< "\n";
答案 1 :(得分：0)
您的评论是正确的。
修订： 创建一个struct s1;
有三个值 - 人物指数，汉堡包数和煎饼数。
填写并将s1＆s放入矢量
在汉堡包计数上使用std :: sort。 - 打印您想要的值。
在煎饼数上使用std :: sort - 打印你想要的值。
较大的计数将在一端（而最小的计数在另一端）。
2015年5月14日修订：
时间回答你的问题 - 如何缩小规模。
您只需识别模式...并将其与for循环匹配。
模式构思1 - personId序列是1到10;
引导我们：
for (int personId = 1; personId <= 10; ++personId)
{
}
对于每个人，您的代码会提示计数 - 煎饼，然后是汉堡包：
for (int personId = 1; personId <= 10; ++personId)
{
cout << "How many pancakes did p" << personId << " eat?" << endl;
int p1 = 0;
cin >> p1;
cout << "How many hamburgers did p" << personId << " eat?" << endl;
int p2 = 0;
cin >> p2;
// perhaps these should be joined into a single prompt to get 2 numbers
// example:
cout << "Enter pancake count, then hamburger count: " << endl;
int p1 = 0;
int p2 = 0;
cin >> p1 >> p2;
// error handling is somewhat easier in the 1st style.
}
要确定哪个是最少或最多，我建议您将这些项目放入矢量中进行排序。
struct S1 {
int personId;
int pancakeCount;
int hamburgerCount;
// define ctor
S1(void) : personId(0), pancakeCount(0), hamburgerCount(0)
{
}
};
vector<S1> s1Vec;
for (int personId = 1; personId <= 10; ++personId)
{
cout << "How many pancakes did p" << personId << " eat?" << endl;
int p1 = 0;
cin >> p1;
cout << "How many hamburgers did p" << personId << " eat?" << endl;
int p2 = 0;
cin >> p2;
{
S1 s1; // declare an instance, and fill in
s1.personId = personId;
s1.pancackecount = p1;
s1.hamburgerCount = p2;
s1Vec.push_back(s1); // default cpy-ctor will capture into vec
} // THIS brace triggers the dtor of s1 ... currently does nothing,
// but sometimes the intermediate structs / classes need cleaning
}
// now you have all ids and counts captured to the vector
// sort s1Vec by pancakeCount, and
// print out s1Vec.begin() and .end()
// sort s1Vec by hamburgerCount, and
// print out s1.begin() and .end()
我建议你为S1添加一个show方法......
struct S1 {
int personId;
int pancakeCount;
int hamburgerCount;
// define ctor to initialize values
S1(void) : personId(0), pancakeCount(0), hamburgerCount(0)
{
}
std::string show(void)
{
stringstream ss;
ss << "person: " << personId
<< " p1: " << pancakeCount
<< " p2: " << hamburgerCount << endl;
return (ss.str());
}
std::string showP1(void)
{
stringstream ss;
cout << "person: " << personId
<< " p1: " << pancakeCount << endl;
return (ss.str());
}
std::string showP2(void)
{
stringstream ss;
cout << "person: " << personId
<< " p2: " << hamburgerCount << endl;
return (ss.str());
}
};
这通过使用循环和结构减少了代码的大小。
您的代码可以被视为已展开的＆＃39;循环，这是您可能考虑的性能问题。但是，嗨，用户I / O（提示和响应）永远不会高性能。
答案 2 :(得分：0)
我不明白为什么答案必须如此复杂。
vector：
auto num_eaten = std::vector<int>(10);
for (auto i = 0u; i < num_eaten.size(); ++i) {
cout << "Enter number of hamburgers eaten by " << (i + 1) << "> ";
cin >> num_eaten[i];
}
auto min_it = std::min_element(num_eaten.begin(), num_eaten.end());
cout << (std::distance(num_eaten.begin(), min_it) + 1) << " ate the least"
<< endl;
auto max_it = std::max_element(num_eaten.begin(), num_eaten.end());
cout << (std::distance(num_eaten.begin(), max_it) + 1) << " ate the most"
<< endl;
或map：
auto eaten = std::unordered_map<int, int>{};
for (auto i = 1; i < 10; ++i) {
cout << "Enter number of hamburgers eaten by " << i << "> ";
cin >> eaten[i];
}
auto min_it = std::min_element(
eaten.begin(), eaten.end(),
[](auto const &lhs, auto const &rhs) { return lhs.second < rhs.second; });
cout << min_it->first << " ate the least: " << min_it->second << endl;
auto max_it = std::min_element(
eaten.begin(), eaten.end(),
[](auto const &lhs, auto const &rhs) { return lhs.second > rhs.second; });
cout << max_it->first << " ate the most: " << max_it->second << endl;