hdu-1695 GCD(莫比乌斯反演)

Posted 2020-08-02 LittlePointer

题目链接：

GCD

Time Limit: 6000/3000 MS (Java/Others)  

  Memory Limit: 32768/32768 K (Java/Others)

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

 

 

题意：

 

求gcd(x,y)==k的(x,y)对数;1<=x<=b,1<=y<=d;

 

思路：

 

gcd(x,y)=k,即gcd(x/k,y/k)=1;

变成求[1,b/k][1,d/k]中的互质的对数;

f(d)表示gcd(x,y)=d的对数;F(d)表示d|gcd(x,y)的对数;

这里求的是f(1);由莫比乌斯反演f(n)=∑(n|d)µ(d/n)F(d)知

f(1)=∑µ(i)*F(i)   (1<=i<=min(b/k,d/k));

 

AC代码：