GCD（关于容斥原理）

Problem Description

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.

Yoiu 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 <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 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

 

这个问题是用容斥原理来解决的，我们知道的是根据gcd(x,y)=k--> gcd(x/k,y/k)=1;

这样子的话，我们当k为0的时候特判一下就行了，剩下的就是根据容斥原理来做了，已知区间[1,y]如何求z和区间里面的几个数互质。。。

但是有一点我是真的不明白，为什么不超时啊，3000组数据的话，我真的不知道为什么不超时，求大神指点啊。