莫比乌斯函数 hdu6053

TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2997    Accepted Submission(s): 1121

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

144 4 4 4

 

Sample Output

Case #1: 17

 

Source

2017 Multi-University Training Contest - Team 2

#include<bits/stdc++.h>using namespace std;#define N 200011const int mod=1e9+7;typedef long long LL;int mu[110000];int MAXN=100000;int a[110000];int num[210000];void Moblus(){    mu[1]=1;    for(int i=1; i<=MAXN; i++)        for(int j=i+i; j<=MAXN; j+=i)            mu[j]-=mu[i];}LL qpow(LL a,LL b){    LL ans=1;    while(b)    {        if(b&1)            ans=(ans*a)%mod;        b>>=1;        a=(a*a)%mod;    }    return ans;}int main(){    Moblus();    int t;    scanf("%d",&t);    for(int k=1; k<=t; k++)    {        int i,j,n;        scanf("%d",&n);        memset(a,0,sizeof(a));        int mn=100005;        int mx=-1;        for(i=0; i<n; i++)        {            int x;            scanf("%d",&x);            mn=min(x,mn);            mx=max(x,mx);            a[x]++;        }        num[0]=a[0];        for(i=1; i<=100100; i++)                num[i]=num[i-1]+a[i];        LL sum=0;        int l,r;        for(i=2; i<=mn; i++)        {            LL ans=1;            for(j=1; j*i<=mx; j++)            {                r=(j+1)*i-1;                l=j*i-1;                if(r>100000)                    r=100000;                ans=(ans*qpow(j,num[r]-num[l]))%mod;            }            sum=(sum-ans*mu[i]%mod+mod)%mod;        }        printf("Case #%d: %lld\n",k,sum);    }    return 0;}