HDU - 1695 GCD —— 莫比乌斯反演

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本文介绍了一种解决特定数学问题的方法——求解在一定范围内满足最大公约数等于给定值的整数对的数量。通过莫比乌斯反演等数学技巧，文章提供了一个高效的算法实现。

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GCD
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15143 Accepted Submission(s): 5793

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).

题意：

求1<=x<=b,1<=y<=d中满足gcd(x,y)=k的对数

思路：

设n=b/k，m=d/k，

要求gcd(x,y)=k，即求满足1<=x<=n,1<=y<=m中gcd(x,y)=1的对数，

设f(d)为满足gcd(x,y)=t的对数，

则可以构造F(d)为满足gcd(x,y)%d=0的对数，那么F(d)=(n/d)*(m/d)，

由莫比乌斯反演，可以通过构造的F函数求出f函数，

因为(a,b)和(b,a)不同时计算，所以要再减去重复的部分。

要注意特判k=0的时候，否则会RE....

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>
#include <vector>
#include <bitset>
#define max_ 1100000
#define inf 0x3f3f3f3f
#define ll long long
#define les 1e-8
#define mod 364875103
using namespace std;
bool vis[max_];
int prime[max_];
int pl=0;
ll mu[max_];
int a,b,c,d,k;
void getmu()
{
    mu[1]=1;
    for(int i=2;i<max_;i++)
    {
        if(vis[i]==false)
        {
            prime[++pl]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=pl&&prime[j]*i<max_;j++)
        {
            vis[prime[j]*i]=true;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            else
            mu[prime[j]*i]=-mu[i];
        }
    }
}
int main(int argc, char const *argv[]) {
    getmu();
    int t;
    scanf("%d",&t);
    for(int r=1;r<=t;r++)
    {
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(k==0)
        {
            printf("Case %d: 0\n",r );
            continue;
        }
        int n=b/k;
        int m=d/k;
        if(n>m)
        swap(n,m);
        ll ans1=0,ans2=0;
        for(int i=1;i<=n;i++)
        {
            ans1+=mu[i]*(n/i)*(m/i);
            ans2+=mu[i]*(n/i)*(n/i);
        }
        printf("Case %d: %lld\n",r,ans1-ans2/2);
    }
    return 0;
}