Question

该程序应该添加1/2 ^ 1 + 1/2 ^ 2 + 1/2 ^ 3 .... 1/2 ^ n（用户输入n次幂）。它应显示分数（1/2 + 1/4 + 1/8 ....）然后找到它们的总和并显示最后的总和（例如：1/2 + 1/4 + 1/8 =。 125）它在用户输入5时有效，但任何其他数字显示错误的总数。我得到的总和大于1，这是不正确的。我该如何解决这个问题？

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

int main()
{
    int denom,              // Denominator of a particular term
        finalDenom,         // Denominator of the final term
        nthTerm;            // Nth term run
    double sum = 0.0;          // Accumulator that adds up all terms in the series
    char repeat;

    do
    {
        cout << "This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n\n";
        cout << "What should n be in the final term (between numbers 2 and 10)? ";
        cin >> finalDenom;

        nthTerm = 0;
        for (denom = 2; nthTerm <= (finalDenom - 1); denom *= 2)
        {
            cout << "1/" << denom;
            ++nthTerm;

            if (denom != finalDenom)
            {
                cout << " + ";
            }
            else if (denom == finalDenom)
            {
                cout << " = ";
            }
            sum += pow(denom, -1);
        }

        cout << sum << endl << endl << endl;

        cout << "Do you wish to compute another series? ";
        cin >> repeat;
        repeat = toupper(repeat);
    } while ((repeat == 'Y'));


    return 0;
}

Answer 1

我认为你的方法效率低且不精确。这是一个几何级数。可以使用单个表达式找到该序列的总和。

在这个公式中用1/2代替r。

Answer 2

您的迭代计数器不是denom，而是nthTerm。因此，您的if else语句应针对finalDenom而非nthTerm检查denom。

此外，您看到的结果大于1，原因是：您可能在继续执行中测试了代码（使用do while循环），而不会在每次执行时将sum变量重置为零。

同样在评论中提到的用户Jean，sum += 1.0/denom;就足够了。这应该有效：

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

int main()
{
    int denom,              // Denominator of a particular term
        finalDenom,         // Denominator of the final term
        nthTerm;            // Nth term run
    double sum = 0.0;          // Accumulator that adds up all terms in the series
    char repeat;

    do
    {
        cout << "This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n\n";
        cout << "What should n be in the final term (between numbers 2 and 10)? ";
        cin >> finalDenom;

        sum = 0;
        nthTerm = 0;

        for (denom = 2; nthTerm < finalDenom; denom *= 2)
        {
            cout << "1/" << denom;
            ++nthTerm;

            if (nthTerm != finalDenom)
                cout << " + ";
            else 
                cout << " = ";

            sum += 1.0 / denom;
        }

        cout << sum << endl << endl;

        cout << "Do you wish to compute another series? ";
        cin >> repeat;

        repeat = toupper(repeat);

    } while (repeat == 'Y');

    return 0;
}

结果：

This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n
What should n be in the final term (between numbers 2 and 10)? 6
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 = 0.984375

Do you wish to compute another series? y
This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n
What should n be in the final term (between numbers 2 and 10)? 5
1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 0.96875

Do you wish to compute another series? y
This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n
What should n be in the final term (between numbers 2 and 10)? 4
1/2 + 1/4 + 1/8 + 1/16 = 0.9375

Do you wish to compute another series? y
This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n
What should n be in the final term (between numbers 2 and 10)? 3
1/2 + 1/4 + 1/8 = 0.875

Do you wish to compute another series? n

另外请注意，您应该添加一些用户输入程序的验证，至少最终用语在[2,10]范围内。

Answer 3

如上所述，你的问题在于传导。但是我会将它完全抛弃。

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

int main()
{
    int denom,              // Denominator of a particular term
        finalDenom,         // Denominator of the final term
        nthTerm;            // Nth term run
    double sum = 0.0;          // Accumulator that adds up all terms in the series
    char repeat;

    do
    {
        cout << "This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n\n";
        cout << "What should n be in the final term (between numbers 2 and 10)? ";
        cin >> finalDenom;
        denom=2;
        for (nthTerm = 0; nthTerm < finalDenom; nthTerm++)

        {
            denom*=2;

            cout << "1/" << denom;
            cout << " + ";                      
            sum += pow(denom, -1);
        }

        cout << " = ";

        cout << sum << endl << endl << endl;

        cout << "Do you wish to compute another series? ";
        cin >> repeat;
        repeat = toupper(repeat);
    } while ((repeat == 'Y'));


    return 0;
}

输出：

This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n
What should n be in the final term (between numbers 2 and 10)? 1
1/4 +  = 0.25


Do you wish to compute another series? y
This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n
What should n be in the final term (between numbers 2 and 10)? 2
1/4 + 1/8 +  = 0.625


Do you wish to compute another series? y
This program sums the series 1/2^1 + 1/2^2 + 1/2^3 + . . . + 1/2^n
What should n be in the final term (between numbers 2 and 10)? 4
1/4 + 1/8 + 1/16 + 1/32 +  = 1.09375


Do you wish to compute another series? n

Exit code: 0 (normal program termination)

Answer 4

我相信它更容易获得值Something。每个循环的一半，或somthing = 1/pow(2,n)（1除以2的幂）。总和为1 - something。更少的代码总是很好。：）

另一项改进是：

for (denom = 2; nthTerm < finalDenom; denom *= 2)
    {
        cout << "1/" << denom;
        ++nthTerm;

            cout << " + ";

        sum += 1.0 / denom;
    }
    Cout << "=";

对此代码进行细微调整

所以我的代码将是

for (denom = 2; nthTerm < finalDenom; denom *= 2)
{
    cout << "1/" << denom;
    ++nthTerm;

        cout << " + ";
}
Cout << "=";
Sum = 1 - (pow(2,finalDenom));

如果您认为这是一个错误的答案，请解释原因，以便我可以改进我的答案。

C ++如何计算运行总分数的总和？

4 个答案: