C++ : Count even/odd numbers in a range

Question

我的程序必须计算一个范围内有多少数字是偶数，有多少是奇数，但我似乎无法弄清楚。它有点管用但是当我把数字放进去时，它就会胡说八道。在编程方面我是一个极端的菜鸟，我认为问题必须在第 21 行 for (i=n; i<=m; i++) { ?但我不确定。我有一本编程书，但用处不大，也许有人可以帮忙？

#include <iostream>

using namespace std;

int main()
{
    int n;
    int m;
    int i;
    int a;
    int b;

    cout << "Enter a number that begins interval: ";
    cin >> n;
    cout << "Enter a number that ends interval: ";
    cin >> m;
    a=0;
    b=0;


    for (i=n; i<=m; i++) {
        if (i%2 == 0){
                a=a+i;
    }
    else {
          b=b+i;
    }
}
    cout << " unequal numbers: " << a << endl;
    cout << "  equal  numbers: " << b << endl;

Answer 1

假设您指的是偶数和奇数，您的问题出在这段代码中:

for (i=n; i<=m; i++) {
    if (i%2 == 0){
        a=a+i;   // increase number of even numbers by i
    }
    else {
        b=b+i;   // increase number of odd numbers by i
    }
}

你可能想要做的是加 1(而不是 i 是什么):

 for (i = n; i <= m; ++i) {
    if (i % 2 == 0)
        ++a;   // increase number of even numbers by one
    else 
        ++b;   // increase number of odd numbers by one
 }

我还建议使用更好的变量名，例如 even 和 odd 而不是 a 和 b 等等。它使每个人(甚至您)都更容易理解代码。

再多一点提示。好的做法是在声明变量后立即分配变量:

int m = 0;

您可以在 for 循环内声明变量，在您的情况下无需在循环外声明它:

for (int i = n; i <= m; ++i) { ... }

示例它如何改变代码的外观和清晰度:

#include <iostream>

using namespace std;

int main() {
    int from = 0,
        to   = 0,
        even = 0,
        odd  = 0;

    cout << "Enter a number that begins interval: ";
    cin >> from;

    cout << "Enter a number that ends interval: ";
    cin >> to;

    for (int i = from; i <= to; ++i) {
        if (i % 2 == 0)
            ++even;
        else 
            ++odd;
    }

    cout << " even numbers: " << even << endl;
    cout << " odd numbers: " << odd << endl;

    return 0;  // don't forget this! main is function returning int so it should return something
}