hdu 1695 GCD（莫比乌斯反演）

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6081    Accepted Submission(s): 2223

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

Hint

For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).

 Source

2008 “Sunline Cup” National Invitational Contest

 

题意：  给你5个整数 a , b , c , d , k ，要你在[a,b]中找到一个x,在[c,d]中找到一个y，使得gcd(x,y)=k(注意 gcd（x,y）=k,与gcd(y,x)=k算作同一种足组合，虽然x,y的取值范围不同)。要你求出总共有多少种组合。

分析：   对于gcd(x,y)=k;

          常规的化简就是 gcd(x/k,y/k)=1;  --->即【a,b】范围缩减到--》【a/k,b/k】，同理[c,d] 缩减到-->[c/k,d/k];

但是这样依旧会tle到天上去......,怎么办？

        还需要优化.  这时候就需要用到容斥原理了。