HDU-1695-GCD(莫比乌斯反演)

最新推荐文章于 2021-11-03 21:35:13 发布

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最新推荐文章于 2021-11-03 21:35:13 发布

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分类专栏：数论

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本文介绍了一种解决特定GCD问题的算法实现，通过预处理得到莫比乌斯函数值，快速计算在给定范围内两数的最大公约数等于指定值的数对数量。文章详细展示了算法流程及核心代码。

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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

思路：GCD(X,Y)=K可转化为GCD(X,Y)=1,对应的上界变成b/k和d/k。

#include <stdio.h>
#include <string.h>

int prime[100005],mu[100005];
bool check[100005];

void Mobius(int n)
{
    int i,j,cnt;

    memset(check,0,sizeof check);

    cnt=0;
    mu[1]=1;

    for(i=2;i<=n;i++)
    {
        if(!check[i])
        {
            prime[cnt++]=i;

            mu[i]=-1;
        }

        for(j=0;j<cnt;j++)
        {
            if(i*prime[j]>n) break;

            check[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,a,b,c,d,k,i,cases;
    long long ans,temp;

    Mobius(100000);

    scanf("%d",&T);

    for(cases=1;cases<=T;cases++)
    {
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);

        if(!k)
        {
            printf("Case %d: 0\n",cases);

            continue;
        }

        if(b>d) b=b+d,d=b-d,b=b-d;//swap(b,d);

        b/=k;
        d/=k;

        ans=temp=0;

        for(i=1;i<=b;i++) ans+=(long long)mu[i]*(b/i)*(d/i);//求出(1,b),(1,d)之间的互质数的对数

        for(i=1;i<=b;i++) temp+=(long long)mu[i]*(b/i)*(b/i);//求出(1,b),(1,b)之间的互质数的对数

        printf("Case %d: %I64d\n",cases,ans-temp/2);
    }
}