HDU 6053 TrickGCD （莫比乌斯函数）

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

题意

给出数组 A ，问有多个种 B 数组满足所给条件。

思路

针对条件，我们可以枚举 gcd ，显然对于每一个素因子 i 在范围 [i,j] 下共有 ji 个数可以整除它，假设 A 中共有 cnt 个数字处于 [j,j+i−1] 这个范围内，这也就相当于有 cnt 个位置的数可以在 ji 个因子中随意变动，共有 (ji)cnt 种方案，而对于每一个因子，其结果为 ∑i|j(ji)cnt 。

最终通过容斥或者莫比乌斯去掉重复部分即可。

AC 代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include<iostream>
using namespace std ;

#define inf 0x3f3f3f
typedef __int64 LL;
const int maxn = 1e5+10;
const int mod = 1e9+7;
LL mu[maxn];
LL sum[maxn<<1];
LL cnt[maxn<<1];

void init()
{
mu[1]=1;
for(int i=1; i<maxn; i++)
for(int j=i+i; j<maxn; j+=i)
mu[j]-=mu[i];
}

LL mult(LL a,LL n)
{
LL res=1;
while(n)
{
if(n&1)res=(res*a)%mod;
a=(a*a)%mod;
n>>=1;
}
return res;
}

int main()
{
ios::sync_with_stdio(false);
init();
int T;
cin>>T;
for(int ti=1; ti<=T; ti++)
{
int n,minn=inf;
memset(cnt,0,sizeof(cnt));
cin>>n;
for(int i=0; i<n; i++)
{
int x;
cin>>x;
cnt[x]++;
minn=min(minn,x);
}
LL ans=0;
for(int i=1; i<maxn*2; i++)
sum[i]=sum[i-1]+cnt[i];
for(int i=2; i<=minn; i++)
{
LL temp=1LL;
if(mu[i])
{
for(int j=i; j<maxn; j+=i)
temp=(temp*mult(j/i,sum[j+i-1]-sum[j-1]))%mod;
}
ans=(ans-temp*mu[i]+mod)%mod;
}
cout<<"Case #"<<ti<<": "<<ans<<endl;
}
return 0;
}