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最新推荐文章于 2020-09-26 17:04:27 发布

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莫比乌斯反演

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本文介绍了一道使用莫比乌斯反演解决的算法题，题目要求计算满足特定条件的不同数组数量。通过实现代码详细展示了如何运用莫比乌斯函数及反演技巧来求解此类问题。

Problem Description
You are given an array A , and Zhu wants to know there are how many different array B satisfy the following conditions?

1≤Bi≤Ai
For each pair( l , r ) (1≤l≤r≤n) , gcd(bl,bl+1…br)≥2

Input
The first line is an integer T(1≤T≤10) describe the number of test cases.

Each test case begins with an integer number n describe the size of array A.

Then a line contains n numbers describe each element of A

You can assume that 1≤n,Ai≤105

Output
For the kth test case , first output “Case #k: ” , then output an integer as answer in a single line . because the answer may be large , so you are only need to output answer mod 109+7

Sample Input
1
4
4 4 4 4

Sample Output
Case #1: 17

Source
2017 Multi-University Training Contest - Team 2

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liuyiding | We have carefully selected several similar problems for you: 6055 6054 6053 6052 6051

莫比乌斯反演的第一次用….
其实就是说莫比乌斯反演那个公式可以反过来用….
用大的算小的…

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=200105;
int normal[maxn];
int mu[maxn];
int prime[maxn];
int pcnt,u=0;
void Init()
{
    memset(normal,0,sizeof(normal));
    mu[1] = 1;
    pcnt = 0;
    for(int i=2; i<maxn; i++)
    {
        if(!normal[i])
        {
            prime[pcnt++] = i;
            mu[i] = -1;
        }
        for(int j=0; j<pcnt&&i*prime[j]<maxn; j++)
        {
            normal[i*prime[j]] = 1;
            if(i%prime[j]) mu[i*prime[j]] = -mu[i];
            else
            {
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}
int tu[maxn],sum[maxn],n,q;
ll mo=1e9+7;
ll ksm(ll ds,ll zs)
{
    ll fs=1;
    while(zs)
    {
        if(zs&1)fs*=ds;

        fs%=mo;
        ds*=ds;
        ds+=mo;
        ds%=mo;
        zs>>=1; //cout<<zs<<endl;
    }
    return (fs+mo)%mo;
}
ll f[maxn],F[maxn];
int main()
{
    Init();
//  cout<<ksm(2,2)<<endl;
//  return 0;
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        memset(f,0,sizeof(f));
        memset(F,0,sizeof(F));
        memset(tu,0,sizeof(tu));
        memset(sum,0,sizeof(sum));
        int shang=0,xia=1e9;
        for(int a=1;a<=n;a++)
        {
            scanf("%d",&q);
            tu[q]++;
            shang=max(shang,q);
            xia=min(xia,q);
        }
        for(int a=1;a<=200010;a++)
        {
            sum[a]=sum[a-1]+tu[a];
        }
        for(int a=2;a<=xia;a++)
        {
            F[a]=1;
            for(int b=a;b<=shang;b+=a)
            {
                int qj=b/a;
        //      int y=min(,shang);
            //  cout<<"ccccc";
    //      cout<<sum[y]-sum[b-1]<<endl<<qj<<endl;
                F[a]=(F[a]*ksm(qj,sum[b+a-1]-sum[b-1])%mo+mo)%mo;
            }//cout<<"asd";
        }
    //  return 0;
        for(int a=2;a<=xia;a++)
        {
            for(int b=1;;b++)
            {
                if(b*a>xia)break;
                f[a]=(f[a]+mu[b]*F[a*b]+mo)%mo;
            }
        }

        ll dan=0;
        for(int a=2;a<=xia;a++)
        {
            dan=(dan+f[a]+mo)%mo;
        }
        printf("Case #%d: %lld\n",++u,dan%mo);
    }
}