给你5个数a，b，c，d，k。x属于[a，b]y属于[c，d]。 问你有多少对（x，y）的公约数为k

题目大意：给你5个数a，b，c，d，k。x属于[a，b]y属于[c，d]。 问你有多少对（x，y）的公约数为k。 注意（x，y）和 （y，x）视为同一对。

x是[1,b]，y是[1,d]，求GCD(x,y)=k的对数（x,y无序）

对x，y都除以k，则求GCD(x,y)=1

此时枚举x，问题转化为[1,d]区间内与x互素的数字个数，这个问题是hdu 4135

有一个特殊的地方是x，y无序，对于这点只要保证x始终小于y就可以了

              注意特判k==0的情况

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). 
    
    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int s[100020],k;
void init(int m)
{
    k=0;
    for(int i=2; i*i<=m; i++)
    {
        if(m%i==0)
        {
            s[k++]=i;
            while(m%i==0)
                m/=i;
        }
    }
    if(m>1)
        s[k++]=m;
}
long long quc(int m)
{
    int t=0,p[100020],z;
    long long sum=0;
    p[t++]=-1;
    for(int i=0; i<k; i++)
    {
        z=t;
        for(int j=0; j<z; j++)
        {
            p[t++]=p[j]*s[i]*(-1);
        }
    }
    for(int i=1; i<t; i++)
    {
        sum+=m/p[i];
    }
    return sum;
}
int main()
{
    int n,a,b,c,d,m,g=0;
    long long sum;
    scanf("%d",&n);
    while(n--)
    {
        sum=0;
        scanf("%d %d %d %d %d",&a,&b,&c,&d,&m);
        g++;
        if(m==0||m>b||m>d)
            {printf("Case %d: 0\n",g);
            continue;
            }
        b=b/m;
        d=d/m;
        if(b>d)
            swap(b,d);
        for(int i=1; i<=b; i++)
        {
            init(i);
            sum+=d-quc(d)-i+1+quc(i-1);
            //printf("%d %d\n",quc(i-1),quc(d));
        }
        printf("Case %d: %I64d\n",g,sum);
    }
}