Question

它应该打印Fibonacci系列直到一个位置，但它只打印1 1 2，即使我要求超过前三个元素。我该如何解决这个问题？

#include <iostream>
using std::cout;
using std::cin;

int main()
{
    cout << "Enter a number: ";
    int number;
    cin >> number;
    int count = 1;
    int a = 1;                  //The first number of the Fibonacci's serie is 1
    int b = 1;                  //The second number of the Fibonacci's serie is 2
    while (count <= number)
    {
        if (count < 3)
            cout << "1 ";
        else
        {
            number = a + b;     //Every number is the sum of the previous two
            cout << number << " ";
            if (count % 2 == 1)
                a = number;
            else
                b = number;
        }
        count++;
    }

    return 0;
}

Answer 1

您在这里使用number作为循环迭代的最大数量：

while (count <= number)

但是然后在循环中你使用与当前Fib值相同的变量来为每次迭代输出。

number = a + b;     //Every number is the sum of the previous two
cout << number << " ";

这导致循环过早终止。您需要为这两个不同的值选择不同的变量名称。

Answer 2

这就像交换变量的值。您使用数字作为限制，但在循环中您使用的是创建逻辑错误的相同变量。做以下更改并完成（Y）。

Answer 3

您可以尝试以下代码：

int main()
{
    int n, t1 = 0, t2 = 1, nextTerm = 0;

    cout << "Enter the number of terms: ";
    cin >> n;

    cout << "Fibonacci Series: ";

    for (int i = 1; i <= n; ++i)
    {
      // Prints the first two terms.
        if(i == 1)
        {
            cout << " " << t1;
            continue;
        }
        if(i == 2)
        {
            cout << t2 << " ";
            continue;
       }
        nextTerm = t1 + t2;
        t1 = t2;
        t2 = nextTerm;

        cout << nextTerm << " ";
    }
    return 0;
}

这段代码有什么问题？斐波纳契系列

3 个答案: