创建一个C++计算奇数、偶数和零数的程序

Create a program that counts number of odd, even and zero numbers C++

本文关键字：程序 计算 C++ 一个 创建      更新时间：2023-10-16
#include <iostream>using namespace std;
const int LIMIT=10;
int main () {
  float counter;
  int number=0;
  int zeros=0;
  int odds=0;
  int evens=0;
  cout << "Please enter " << LIMIT << " integers, " << "positive, negative, or zeros." << endl;
  cout << "The numbers you entered are:" << endl;
  for (counter=1;      counter <=LIMIT;      counter++) {
    cin>>number;
    switch (number / 2) {
      case 0: evens++;
      if (number=0) zeros++;
      case 1: case -1: odds++;
    }
  }
  cout << endl;
  cout << "There are " << evens << " evens, " << "which includes " << zeros << " zeros." << endl;
  cout << "The number of odd numbers is: " << odds << endl;
  return 0;
}

大家好

我有一个校队问题，让我一整天都难倒。我需要修改上面的脚本以允许我输入 10 个可变整数，程序必须返回偶数总数、奇数总数和零总数。

我已经尝试了多种解决方案，包括(数字 % 2 == 0(，以使我的情况在我的开关参数下工作，但我缺少一些东西。

请有人帮助我走上正确的道路。

(我知道我需要删除负面情况，但我想发布原始代码，以防我取出某些东西或修复需要的东西(

#include <iostream>using namespace std;
const int LIMIT=10;
int main () {
  float counter;
  int number=0;
  int zeros=0;
  int odds=0;
  int evens=0;
  int n=0;
  cout << "Please enter " << LIMIT << " integers, " << "positive, negative, or zeros." << endl;
  cout << "The numbers you entered are:" << endl;
  for (counter=1;
  counter <=LIMIT;
  counter++) {
    cin>>number;
    n=(number%2);
    switch (n) {
      case 0: evens++;
      if (number==0) zeros++;
      break;
      case 1: odds++;
    }
  }
  cout << endl;
  cout << "There are " << evens << " evens, " << "which includes " << zeros << " zeros." << endl;
  cout << "The number of odd numbers is: " << odds << endl;
  return 0;
}

多亏了帕帕加加设法让它工作 - 我想我太累了:)

std::vector<int> myvec = {0, 2, 3, 5, 0, 0, 5, 4, 9};
std::vector<int> zeros;
std::vector<int> evens;
std::vector<int> odds;
for (auto i : myvec) {
    if (i % 2 == 0) {
        if (i == 0) zeros.push_back(i);
        else evens.push_back(i);
    }
    else
        odds.push_back(i);
}
std::cout << zeros.size();
std::cout << evens.size();
std::cout << odds.size();
return 0;
}

如果您不希望零在偶数向量中，这将起作用。

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