HDU 1695 GCD

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.

题目大意:
求出[1,b],[1,d]中gcd为k的数对的个数
解题报告:
比较简单,转化为求[1,b/k],[1,d/k]中互质的数对个数，就是莫比乌斯反演的板子题了.
设 \(F_i\)表示gcd(a,b)i的数对个数 \(G_i\)表示gcd(a,b)%i0的数对个数,那么答案就是F[1]
显然
\(G_n=(a/n)*(b/n)\)
\(F[1]=\sum_{i=1}^{max(a,b)}\mu_iG_i\)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
const int N=100005;
int CNT=0,num=0,phi[N],prime[N];bool vis[N];
void prework(){
	phi[1]=1;
	for(int i=2;i<N;i++){
		if(!vis[i]){
			prime[++num]=i;
			phi[i]=-1;
		}
		int to;
		for(int j=1;j<=num && prime[j]*i<N;j++){
			to=i*prime[j];
			vis[to]=true;
			if(i%prime[j])
				phi[to]=-phi[i];
			else{
				phi[to]=0;
				break;
			}
		}
	}
}
void work()
{
	int a,b,c,d,k;
	scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
	if(k==0){
		printf("Case %d: 0\n",CNT);
		return ;
	}
	b/=k;d/=k;
	if(b>d)swap(b,d);
	ll ans=0,cnt=0;
	for(int i=1;i<=b;i++){
		ans+=(ll)phi[i]*(b/i)*(d/i);
	}
	for(int i=1;i<=b;i++){
		cnt+=(ll)phi[i]*(b/i)*(b/i);
		//减去重复部分
	}
	ans-=cnt>>1;
	printf("Case %d: %lld\n",CNT,ans);
}

int main()
{
	int T;cin>>T;
	prework();
	while(T--)
		CNT++,work();
	return 0;
}