我应该如何去了解这个递归没有试错解 [英] How should I go about solving this recursion without trial and error

问题描述

int sum_down(int x)
{
    if (x >= 0)
    {
        x = x - 1;
        int y = x + sum_down(x);
        return y + sum_down(x);
    }
    else
    {
        return 1;
    }
}

是什么参数x的这个最小整数值，以使返回的值大于1,000,000？

What is this smallest integer value of the parameter x, so that the returned value is greater than 1.000.000 ?

现在我只是在做它的试验和错误，并因为这个问题是通过纸张格式要求。我不认为我有足够的时间做试验和错误。问题是，你们如何想象这种快，使得它可以轻松解决。谢谢你们，我是新来编程，以便在此先感谢！

Right now I am just doing it by trial and error and since this question is asked via a paper format. I don't think I will have enough time to do trial and error. Question is, how do you guys visualise this quickly such that it can be solved easily. Thanks guys and I am new to programming so thanks in advance!

推荐答案

递归的逻辑：

x = x - 1;
int y = x + sum_down(x);
return y + sum_down(x);

可以简化为：

can be simplified to:

x = x - 1;
int y = x + sum_down(x) + sum_down(x);
return y;

其可以简化为：

which can be simplified to:

int y = (x-1) + sum_down(x-1) + sum_down(x-1);
return y;

其可以简化为：

which can be simplified to:

return (x-1) + 2*sum_down(x-1);

将在数学形式，

Put in mathematical form,

f(N) = (N-1) + 2*f(N-1)

与递归终止时， N 是 1 。 F（-1） = 1 。

因此​​，

f(0) = -1 + 2*1 = 1
f(1) =  0 + 2*1 = 2
f(2) =  1 + 2*2 = 5

...

f(18) = 17 + 2*f(17) = 524269
f(19) = 18 + 2*524269 = 1048556

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