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我有几个问题需要解决。这个程序会找到用户想要的数字的平均值。例如,如果他们输入 3,然后输入 1、2、3,则平均值为 2。首先是我的程序如果输入零,则应该停止。所以如果我输入 3 作为将被平均的整数数,然后我输入 1,2,0 它应该停止并且不计算平均值。另外,我需要考虑负数。

#include <iostream>
using namespace std;
double avgVal(int, int);
int main()
{
    int amountOfCases;
    cin >> amountOfCases;
    int * numbers = new int[amountOfCases];
    int sum = 0;
    while(numbers !=0)
    {
    for (int i = 0; i<amountOfCases; i++)
    {
        cin >> numbers[i];
        sum = sum + numbers[i];
    }


    cout<<avgVal(sum, amountOfCases)<<endl;
    delete[] numbers;
    }
    system("pause");
    return 0;
}

double avgVal(int sum, int amountOfCases)
{
    return sum / (double)amountOfCases;
}
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1 回答 1

0

不确定添加while循环时的想法。鉴于您要解决的问题,这不是必需的。

更改行:

while(numbers !=0)
{
for (int i = 0; i<amountOfCases; i++)
{
    cin >> numbers[i];
    sum = sum + numbers[i];
}

cout<<avgVal(sum, amountOfCases)<<endl;
delete[] numbers;
}

for (int i = 0; i<amountOfCases; i++)
{
   cin >> numbers[i];
   if ( numbers[i] == 0 )
   {
      exit(1); // Exit with a non-zero status, indicating failure.
   }
   sum = sum + numbers[i];
}

cout<<avgVal(sum, amountOfCases)<<endl;
delete[] numbers;
于 2014-09-12T03:08:32.600 回答