---

Problem 809

---

Rational Recurrence Relation

The following is a function defined for all positive rational values of $x$.
$$f(x)=\{\begin{array}{ll}x& x\text{ is integral}\\ f(\frac{1}{1-x})& x<1\\ f(\frac{1}{\lceil x\rceil -x}-1+f(x-1))& \text{otherwise}\end{array}$$

For example, $f(3/2)=3$, $f(1/6)=65533$ and $f(13/10)=7625597484985$.

Find $f(22/7)$. Give your answer modulo ${10}^{15}$.

---

有理数递归关系式

如下是定义在所有正有理数$x$上的函数：
$$f(x)=\{\begin{array}{ll}x& \text{若}x\text{是整数}\\ f(\frac{1}{1-x})& \text{若}x<1\\ f(\frac{1}{\lceil x\rceil -x}-1+f(x-1))& \text{其余情况}\end{array}$$

例如，$f(3/2)=3$，$f(1/6)=65533$，$f(13/10)=7625597484985$。

求$f(22/7)$，并将你的答案对${10}^{15}$取余。

---