E - Sum of gcd of Tuples (Hard)

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $500$points

PS:这类%mod的题似乎是只要超出mod都可以%mod，由题意知如果一个个算肯定会超时，所以要从结果为k的gcd的个数切入，另外注意mod的使用，如果答案已为负数记得加一个mod。

Problem Statement

Consider sequences $\{A1,...,AN\}$of length $\mathrm{NN}$ consisting of integers between $1$ and $K$ (inclusive).

There are ${\mathrm{K^N}}^{}$such sequences. Find the sum of $gcd(A1,...,AN)$ over all of them.

Since this sum can be enormous, print the value modulo $(109+7)$.

Here $gcd(A1,...,AN)$denotes the greatest common divisor of ${\mathrm{A1,...,AN}}^{}$.

Constraints

• $\mathrm{2\le N\le 105}$
• $\mathrm{1\le K\le 105}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$$K$

Output

Print the sum of $gcd(A1,...,AN)$over all ${\mathrm{KN}}^{}$sequences, modulo $(109+7)$.

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Sample Input 1 Copy

Copy

3 2

Sample Output 1 Copy

Copy

9

$gcd(1,1,1)+gcd(1,1,2)+gcd(1,2,1)+gcd(1,2,2)$$+gcd(2,1,1)+gcd(2,1,2)+gcd(2,2,1)+gcd(2,2,2)$$=1+1+1+1+1+1+1+2=9$

Thus, the answer is $9$.

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Sample Input 2 Copy

Copy

3 200

Sample Output 2 Copy

Copy

10813692

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Sample Input 3 Copy

Copy

100000 100000

Sample Output 3 Copy

Copy

742202979

Be sure to print the sum modulo $(109+7)$.

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
const int N=1e5+5;
ll cnt[N];
ll n,k,ans;

ll pow(ll num){
    ll res=1,m=n;
while(m>0){
    if(m&1)
    res=res*num%mod;
    num=num*num%mod;
    m>>=1;
}
//cout<<res<<'.'<<endl;
return res;
}

int main(){
ios::sync_with_stdio(0);
cin>>n>>k;
for(ll i=k;i>=1;i--){
    cnt[i]=pow(k/i);
    for(ll j=2;j*i<=k;j++){
        cnt[i]-=cnt[j*i];
    }
    ans+=(mod+(cnt[i]*i)%mod)%mod;
    ans=(ans+mod)%mod;
    //cout<<ans<<endl;
}
cout<<ans<<endl;
return 0;
}