HDU 1695 GCD(オイラー関数、反発原理)

GCD
Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9046    Accepted Submission(s): 3351
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
    
    
    
    
 
    
    
    
    
   x 1~n  ，y 1~m   gcd(x,y)=k;
    
    
    
    
     x 1~n/k  ，y 1~m/k   gcd(x,y)=1; x,y     
    
    
    
    
      ，    n   n     。
    
    
    
    
        1~    ， n     。
    
    
    
    
            ，                  。
    
    
    
    
        ，            ，        ，       ，        。
    
    
    
    
 
    
    
    
    
      ，          ，           ，       
    
    
    
    
 
    
    
    
    
    
    
    
    
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <bitset>

using namespace std;
typedef long long int LL;
#define MAX 1000000
bool check[MAX+5];
LL fai[MAX+5];
LL prime[MAX+5];
LL sprime[MAX+5];
LL q[MAX+5];
int cnt;
void eular()//         
{
	memset(check,false,sizeof(check));
	fai[1]=1;
	int tot=0;
	for(int i=2;i<=MAX+5;i++)
	{
		if(!check[i])
		{
            prime[tot++]=i;
			fai[i]=i-1;
		}
		for(int j=0;j<tot;j++)
		{
			if(i*prime[j]>MAX+5) break;
			check[i*prime[j]]=true;
			if(i%prime[j]==0)
			{
				fai[i*prime[j]]=fai[i]*prime[j];
				break;
			}
			else
			{
				fai[i*prime[j]]=fai[i]*(prime[j]-1);
			}
		}
	}
}
void Divide(LL n)//     
{
    cnt=0;
    LL t=(LL)sqrt(1.0*n);
    for(LL i=0; prime[i]<=t; i++) {
        if(n%prime[i]==0) {
            sprime[cnt++]=prime[i];
            while(n%prime[i]==0)
                n/=prime[i];
        }
    }
    if(n>1)
        sprime[cnt++]=n;
}
LL Ex(LL n)//         
{
    
    LL sum=0;
    LL t=1;
    q[0]=-1;
    for(LL i=0; i<cnt; i++) {
        LL x=t;
        for(LL j=0; j<x; j++){
            q[t]=q[j]*sprime[i]*(-1);
            t++;
        }
    }
    for(LL i=1; i<t; i++)
        sum+=n/q[i];
    return sum;
}
int main()
{
	int t;
	scanf("%d",&t);
	eular();
	int cas=0;
	int a,b,c,d,k;
	while(t--)
	{
		scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
		if(k==0||k>b||k>d)
		{
			printf("Case %d: 0
",++cas);
			continue;
		}
		if(b>d) swap(b,d);
		b/=k;d/=k;
		LL ans=0;
		for(int i=1;i<=b;i++)
             ans+=fai[i];
		for(int i=b+1;i<=d;i++)
		{ Divide(i);ans+=(b-Ex(b));}
		printf("Case %d: %lld
",++cas,ans);
	}
	return 0;

}