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.

 

 


 

 

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).

 

 

Sample Input

 

 

 

 

//分析：题目是要求a<=x<=b,c<=y<=d(其中a,c题目中已经说了必为1)，中有多少对(x,y)（无序的）满足GCD（x,y）=k，那么可以转化成求（x,y）分别在区间1<=x<=b/k,1<=y<=d/k中有多少对满足GCD(x,y)=1； 转换b/=k,d/=k，之后，假设b<=d，那么我们可以分两步来求值：

1、求[1,b]与[1,b]之间有多少对数互质，显然就是phi[1]+phi[2]+...phi[b]即可，用线性求欧拉函数即可！

2\求[1,b]与[b+1,d]之间有多少对数互质，使用容斥定理

两步求值之和即为题意所求！

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=100000;
int q[N+5];
long long p[N+5];
int f(int a,int n)
{
    int w[110],l=0;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            w[++l]=i;
            while(n%i==0)n/=i;
        }
        if(n==1)break;
    }
    if(n!=1)w[++l]=n;
    int sum=0;
    for(int i=1;i< (1<<l);i++)
    {
        int flat=0,s=1;
        for(int j=0;j<l;j++)

            if(i&(1<<j))
            {flat=!flat;
             s*=w[j+1];
            }
            s=a/s;
        if(flat)
            sum+=s;
        else
            sum-=s;
    }
    return (a-sum);
}
int main()
{
    memset(q,0,sizeof(q));
    q[0]=q[1]=1;
    int n=0;
    for(int i=2;i<=N;i++)
    {
        if(!q[i])
        {

            for(int j=i;j<=N;j+=i)
            {
                if(!q[j])q[j]=j;
                q[j]=q[j]/i*(i-1);
            }
        }
    }
    p[1]=1;
    for(int i=2;i<=N;i++)
    {
        p[i]=p[i-1]+q[i];
    }
    int m,a,b,c,d,k;
    int t,kase=0;
    scanf("%d",&t);
    while(t--)
    {
        long long ans;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(k==0||b<k||d<k)
            ans=0;
        else
        {
         b/=k;d/=k;
          if(b>d)swap(b,d);
             ans=0;
            for(int i=b+1;i<=d;i++)
                {int k=f(b,i);
                //cout<<i<<"  * "<<k<<endl;
                 ans+=k;
                }
            ans+=p[b];
        }
           printf("Case %d: %lld\n",++kase,ans);
    }
    return 0;
}