hdu 1695 GCD 数论教学题

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.

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 <= 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).

• 莫比乌斯反演，可比容斥原理节省工作量。
• 所需寻找的二元组中i==j时，仅算1对。

import java.io.*;
import java.util.*;
public class Main {
    final static int maxn=100010;
    static int[] pri,phi,fstp,miu;
    static void get_prime(){
        pri=new int[maxn];
        fstp=new int[maxn];
        phi=new int[maxn];
        miu=new int[maxn];
        phi[1]=1;
        miu[1]=1;
        for(int i=2;i0){
            in.nextInt();
            int n=in.nextInt();
            in.nextInt();
            int m=in.nextInt();
            int k=in.nextInt();
            if (k==0){
                out.println("Case "+(++cas)+": 0");
                continue;
            }
            n/=k;m/=k;
            if (n>m){
                int t=n;n=m;m=t;
            }
            long ans=0;
            for(int i=1;i<=n;i++)
                ans-=(long)miu[i]*(n/i)*(n/i);
            ans/=2;
            for(int i=1;i<=n;i++)
                ans+=(long)miu[i]*(n/i)*(m/i);
            out.println("Case "+(++cas)+": "+ans);
        }
        out.flush();
        out.close();
    }

}