我需要我的程序说“这个值是无效的” [英] I need my program to say "this value is not valid"

问题描述

这是我的工作问题。

三角形数字序列是通过添加自然数生成的。因此，第7个三角形数字将是 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 。前十个术语是： 1,3,6,10,15,21,28,36,45,55，...

The sequence of triangle numbers is generated by adding the natural numbers. Hence, the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

让我们列出前七个三角形数字的因子：

Let us list the factors of the first seven triangle numbers:

1: 1

3: 1,3

6: 1,2,3,6

10: 1,2,5,10

15: 1,3,5,15

21: 1,3,7,21

28: 1,2,4,7,14,28

我们可以看到，28是第一个三角形数字超过五个除数。第一个三角形数字的值超过一百个除数是什么？

We can see that 28 is the first triangle number to have over five divisors. What is the value of the first triangle number to have over one hundred divisors?

这是我的代码。我有错误处理的麻烦。唯一有效的输入是数字。如果我输入一个数字以外的任何东西，我想让它响应错误或类似的东西。

and this this is my code. I am having trouble with the error handling.The only valid inputs are numbers. If I enter anything other than a number, i want it to respond "ERROR" or something like that.

int main()
{

    int div=0,sum=0,num,i=1,count=0,a;
    cout<<"Enter the number of divisors"<< endl;
    cin>>a;
    while(div<=a) { 
        div=0;
        sum=sum+i;
        for(int j=1;j<=sum;j++)
            if (sum%j==0)
                div++;
        chk++;
        i++;
    }
    cout<<"Value of first triangle number is "<<sum<<endl;
    cout<<"Value of triangle number is "<<count<<endl;
    return 0;
}

我的程序在我的伪代码和流程图之前，所以如果any1可以帮助我通过我的伪码n流程图，这将是值得赞赏的。 =）

also i did my program before my pseudocode and the flowchart,so if any1 can help me guide through my pseudocode n flowchart,it would be appreciated.thank you. =)

推荐答案

检查输入和转换是否成功的最简单的方法是检查流是否仍然良好： / p>

The simplest way to check whether the input and conversion succeeded is to check that the stream is still good afterwards:

if (std::cin >> a) {
    // Success
} else {
    // Failure
}

如果要恢复，那么您将需要清除流的状态，并且还要删除错误的输入，然后重试;例如

If you want to recover and try again, then you will need to clear the stream's status, and also remove the bad input, before retrying; something like

std::cin.clear();
std::cin.ignore(unsigned(-1), '\n');

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