Question

我想编写一个只接受奇数的程序，如果输入0，它将输出加法和平均值，而不会将任何偶数值加到平均值和加法中。我坚持不让它采取均衡的价值观.. 到目前为止，我的代码是：

    int num = 0;
    int addition = 0;
    int numberOfInputs = 0;

    cout << "Enter your numbers (only odd numbers), the program will continue asking for numbers until you input 0.." << endl;

    for (; ;) {

    cin >> num;
    numberOfInputs++;
    addition = addition + num;

    if (num % 2 != 0) {
     //my issue is with this part
        cout << "ignored" << endl;
    }

    if (num == 0) {
        cout << "Addition: " << addition << endl;
        cout << "Average: " << addition / numberOfInputs << endl;
        }
}

Answer 1

您的代码解决方案：

由于以下原因，您的代码无效：

问题1：您添加输入数字而不检查是否是偶数

问题2：如果想跳过，那么你的情况应该如下循环：

if (num%2==0) {
cout << "ignored:" <<num << endl;
continue;
}

解决您的问题，我已按以下方式更新您的计划：

#include <iostream>
#include <string>
using namespace std;

int main()
{
  int num = 0;
int addition = 0;
int numberOfInputs = 0;

cout << "Enter your numbers (only odd numbers), the program will continue asking for numbers until you input 0.." << endl;

for (; ;) {

cin>> num;

 if (num%2==0) {
    cout << "ignored:" <<num << endl;
    continue;
}
numberOfInputs++;
addition = addition + num;



if (num == 0) {
    cout << "Addition: " << addition << endl;
    cout << "Average: " << addition / numberOfInputs << endl;
    break;
    }
 } 

}

Answer 2

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

int main() {
    int number;
    int sum=0;
    int average=0;
    int inputArray[20]; // will take only 20 inputs at a time
    int i,index = 0;
    int size;

    do{
    cout<<"Enter number\n";
    cin>>number;

    if(number==0){

        for(i=0;i<index;i++){
            sum = sum + inputArray[i];
        }

        cout << sum;
        average = sum / index;
        cout << average;

    } else if(number % 2 != 0){
               inputArray[index++] = number;
    } else 
    cout<<"skip";

    }
    while(number!=0);
    return 0;
}

您可以在https://www.codechef.com/ide运行并检查此代码通过提供自定义输入

没有接受输入

2 个答案: