HDU 1695 GCD（莫比乌斯反演入门）

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#莫比乌斯反演

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本文介绍了一道关于最大公约数(GCD)的问题及其算法解决方案。问题要求找出特定区间内两数的最大公约数等于给定值的所有数对组合，并计算这些组合的数量。文中提供了一个高效的C++实现代码，通过质数筛法预处理数据来加速计算。

GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15624 Accepted Submission(s): 6013

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).

Source

2008 “Sunline Cup” National Invitational Contest

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#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int a,b,c,d,k,kase;
long long ans1,ans2;
int prime[maxn],vis[maxn],mu[maxn];
void init()
{
    int n=maxn,cnt=0;
    memset(prime,0,sizeof(prime));
    memset(vis,0,sizeof(vis));
    memset(mu,0,sizeof(mu));
    mu[1]=1;
    for(int i=2;i<=n;i++)
    {
        if(!vis[i])
        {
            prime[cnt++]=i;
            mu[i]=-1;
        }
        for(int j=0;j<cnt&&i*prime[j]<n;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]) mu[i*prime[j]]=-mu[i];
            else
            {
                mu[i*prime[j]]=0;
                break;
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    init();
    kase=0;
    while(t--)
    {
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(!k)
        {
            printf("Case %d: 0\n",++kase);
            continue;
        }
        b/=k,d/=k;
        if(b>d) swap(b,d);
        ans1=ans2=0;
        for(int i=1;i<=b;i++)
        {
            ans1+=1LL*mu[i]*(b/i)*(d/i);
            ans2+=1LL*mu[i]*(b/i)*(b/i);
        }
        printf("Case %d: %lld\n",++kase,ans1-ans2/2);
    }
    return 0;
}