HDU 6053 TrickGCD（莫比乌斯反演）

TrickGCD

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

题意：给出长度为n的a数列，求满足条件的B数组的个数，条件：①1<=b[i]<=a[i] ②区间整体gcd>=2

题解：

令g(d)为gcd为d的倍数的答案



根据容斥原理，去掉重复的情况（如g(2)+g(3)-g(6)）





f(i,d)表示序列a中a/d=i的元素个数，可以通过预处理出前缀和+快速幂，之后利用预处理出莫比乌斯函数后进行简单的反演即可算出答案，总复杂度nlogn^2

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<math.h>
#include<queue>
#define INF 1e9
#define ll long long
#define maxn 300005
#include<set>
#include<stack>
using namespace std;
int n,m,mm;
int a[100005];
int mo[100005];
bool isprime[100005];
int num[100005];
ll mod=1e9+7;
void mobi()
{
memset(isprime,true,sizeof(isprime));
isprime[1]=false;
for(int i=2;i<100005;i++)
if(isprime[i])
{
mo[i]=-1;
for(int j=2;i*j<100005;j++)
{
if(j%i==0) mo[i*j]=0;
else mo[i*j]=-mo[j];
isprime[i*j]=false;
}
}
}
ll quickpow(ll n,ll m)
{
ll ans=1;
n%=mod;
while(m)
{
if(m%2==1)
{
ans=ans*n;
ans=ans%mod;
m=m-1;
}
else
{
n=(n*n)%mod;
n%=mod;
m=m/2;
}
}
return ans%mod;
}
int main()
{
int t;
scanf("%d",&t);
mobi();
for(int w=1;w<=t;w++)
{
scanf("%d",&n);
m=0;
mm=100005;
memset(num,0,sizeof(num));
memset(a,0,sizeof(a));
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
m=max(a[i],m);
mm=min(mm,a[i]);
num[a[i]]++;
}
cout<<"Case #"<<w<<": ";
for(int i=1;i<=m;i++)
num[i]+=num[i-1];
ll ans=0;
for(int i=2;i<=mm;i++)
{
if(mo[i]==0) continue;
ll sum=1;
for(int j=1;j<=m/i;j++)
{
ll mi=1ll*(num[min(m,j*i+i-1)]-num[j*i-1]);
sum*=quickpow(1ll*j,mi);
sum%=mod;
}
ans+=sum*(-mo[i]);
ans=(ans+mod)%mod;
}
cout<<ans%mod<<endl;

}
return 0;
}