(HDU 1695)GCD(容斥+莫比乌斯反演)

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13721    Accepted Submission(s): 5233

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).

 

莫比乌斯反演两个公式：

（图片转侵删）

若x和y的gcd为k，那么x/k和y/k是互质的，即gcd（x/k,y/k）==1；

题目就转化成了在【1，b/k】和【1，d/k】的区间内求互质的数的对数；

用到莫比乌斯反演函数在于巧妙的构造函数F（x）和f（x）

在这里设F（x）代表在这两个区间内gcd为x或者x的倍数的数的对数

　　　　 f （x）代表在这两个区间内gcd为         x            的数的对数

那么这两个函数就满足了关系：

 

那么显然答案就是f（1）；

F（x）很容易求出，F（x）=（b/k/x）*（d/k/x）；（第一个区间中x倍数的个数*第二个区间中x倍数的个数）

用推导公式导出关系

f（1）=mu【1】*F(1)+mu【2】*F(2)+。。。。+mu【min（b，d）】*F(min（b，d）)；

因为在题目中(a,b)和(b,a)是一种情况，所以要用最后的结果减去b=d=min（b，d）的情况除以2;

总算搞明白了一点，哭。

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
#include<map>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#define LL long long
#define mod 1000000007
#define inf 0x3f3f3f3f
#define maxn 100010

using namespace std;

LL mu[100100];

void getMu(){
    int N=maxn;
    for(int i=1;i<N;++i){
        int target=i==1?1:0;
        int delta=target-mu[i];
        mu[i]=delta;
        for(int j=2*i;j<N;j+=i)
            mu[j]+=delta;
    }
}

int main()
{
    memset(mu,0,sizeof(mu));
    getMu();
    int t;
    scanf("%d",&t);
    for(int tt=1;tt<=t;tt++)
    {
        LL a,b,c,d,k;
        scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&k);
        if(k==0)
        {
            printf("Case %d: 0
",tt);
            continue;
        }
        b/=k;
        d/=k;
        LL p=0,q=0;
        for(LL i=1;i<=min(b,d);i++)
            p+=(LL)(mu[i]*(b/i)*(d/i));
        for(LL i=1;i<=min(b,d);i++)
            q+=(LL)(mu[i]*(min(b,d)/i)*(min(b,d)/i));
        printf("Case %d: %lld
",tt,p-q/2);
    }
    return 0;
}