HDU1695 GCD

题目重现

Given 5 integers: $a, b, c, d, k$, you’re to find $x \in [a, b], y \in [c, d]$ that $gcd(x, y) = k$. $gcd(x, y)$ means the greatest common divisor of $x$ and $y$.
Since the number of choices may be very large, you’re only required to output the total number of different number pairs.
Please notice that, $(x=5, y=7)$ and $(x=7, y=5)$ are considered to be the same.

You can assume that $a = c = 1$ in all test cases.

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers:
$a, b, c, d, k, 0 < a \le b \le 100,000, 0 < c \le d \le 100,000, 0 \le k \le 100,000$
,as described above.

Output

For each test case, print the number of choices. Use the format in the example.

Sample Input

2
1 3 1 5 1
1 11014 1 14409 9

Sample Output

Case 1: 9
Case 2: 736427

题解

问题基于一个简单的结论：

$$\sum_{1 \le x \le A, 1 \le y \le B} [K | gcd(x, y)] = \lfloor \frac{A}{K} \rfloor \lfloor \frac{B}{K} \rfloor$$

设

$$G_{AB}(K) = \sum_{1 \le x \le A, 1 \le y \le B} [K | gcd(x, y)] \\ F_{AB}(K) = \sum_{1 \le x \le A, 1 \le y \le B} [K = gcd(x, y)] \\ $$

可见有：

$$G_{AB}(K) = \sum_{K | k} F_{AB}(k)$$

由莫比乌斯反演原理有：

$$F_{AB}(k) = \sum_{d \ge 1} {\mu(d) G_{AB}(d k)}$$

所以：

$$F_{AB}(K) = \sum_{1 \le x \le A, 1 \le y \le B} [K = gcd(x, y)] = \sum_{d \ge 1} \mu(d) \lfloor \frac{A}{d K} \rfloor \lfloor \frac{B}{d K} \rfloor$$

此题中要求$(x, y), (y, x)$ 不重复计数

不妨设$A \le B$

$$Ans(A, B, K) = F_{AB}(K) - \frac{F_{AA}(K)}{2}$$

参考代码

• HDU 1695 莫比乌斯反演