bookshelf

Problem Description

Patrick Star bought a bookshelf, he named it ZYG !!

Patrick Star has N book .

The ZYG has K layers (count from 1 to K) and there is no limit on the capacity of each layer !

Now Patrick want to put all N books on ZYG :

1. Assume that the i-th layer has cnti(0≤cnti≤N) books finally.

2. Assume that f[i] is the i-th fibonacci number (f[0]=0,f[1]=1,f[2]=1,f[i]=f[i−2]+f[i−1]).

3. Define the stable value of i-th layers stablei=f[cnti].

4. Define the beauty value of i-th layers beautyi=2stablei−1.

5. Define the whole beauty value of ZYG score=gcd(beauty1,beauty2,...,beautyk)(Note: gcd(0,x)=x).

Patrick Star wants to know the expected value of score if Patrick choose a distribute method randomly !

Input

The first line contain a integer T (no morn than 10), the following is T test case, for each test case :

Each line contains contains three integer n,k(0<n,k≤106).

Output

For each test case, output the answer as a value of a rational number modulo 109+7.

Formally, it is guaranteed that under given constraints the probability is always a rational number pq (p and q are integer and coprime, q is positive), such that q is not divisible by 109+7. Output such integer a between 0 and 109+6 that p−aq is divisible by 109+7.

Sample Input

1 6 8

Sample Output

797202805

这道题还是蛮好的，把欧拉降幂+容斥定理（莫比乌斯）+数论知识

题意：告诉我们有K个桶，一共n个球，把这n个球放到这k 个桶中，然后定义一个魅力值，让我们求期望。魅力值是

所有层的gcd(), 然后魅力值就是 $2^{Fib[gcd]-1}$

前置技能：1. $\bg_white gcd(x^{a}-1,x^{b}-1)=x^{gcd(a,b)}-1$

2. $gcd(Fib[a],Fib[b])=Fib[gcd(a,b)]$

上述两式都可以推广到n个

首先考虑两个桶里面的球，分别为 $a$ 和 $b$ ,对于这两层的魅力值为 $gcd(2^{Fib[a]}-1,2^{Fib[b]}-1)$ ,也就是 $2^{gcd(Fib[a],Fib[b])}-1=2^{Fib[gcd(a,b)]}-1$ ; 所以推广到k个桶，就是 $\bg_white 2^{Fib[gcd(a1,a2,...,ak)]}-1$ , 枚举所有的方案是不可能的，所以我们考虑枚举这个 $\bg_white gcd(a1,a2,...,ak)=p$ , $p$ 的取指最多只有n个，我们是存在枚举的可能的。

直接考虑 $g=p$ ,不好考虑，我们可以考虑 $P$ 的倍数，然后容斥定理计算一下

然后就是莫比乌斯反演的基本操作了

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
const int mod=1e9+7;
const int inv_mod=1e9+6;
const int N=2e6+10;

//begin
int prime[N],mu[N];
bool check[N];
void Mobius(int N){
    mu[1] = 1;
    int tot = 0;
    for(int i=2;i<N;i++){
        if(!check[i]){
            prime[tot ++] = i;
            mu[i] = -1;
        }
        for(int j=0;j<tot;j++){
            if(1LL*i * prime[j] >=1LL*N)break;
            check[1ll*i * prime[j]] = true;
            if(i % prime[j] == 0){
                mu[i * prime[j]] = 0;
                break;
            }else{
                mu[i * prime[j]] = -mu[i];
            }
        }
    }
}
//end

//对于n和m比较小的情况下,C(n,m)的复杂度可以预处理到O(1)
//begin
LL pow_mod(LL base,int n,int mod){
	LL ans=1;
	while(n){
		if(n&1)ans=(ans*base)%mod;
		base=(base*base)%mod;
		n>>=1;
	}
	return ans%mod;
}

int F[2*N],inv_F[2*N];
void init_C_small(int N){
    F[0]=1;
    rep(i,1,N) F[i]=1LL*i*F[i-1]%mod;
    inv_F[N-1]=pow_mod(F[N-1],mod-2,mod);
    per(i,0,N-1)inv_F[i]=1LL*inv_F[i+1]*(i+1)%mod;//invF[i]*i!=1,invF[i+1]*i!*(i+1)=1
}
LL C(int n, int m) {
    if(n < 0 || m < 0 || m > n) return 0;
    if(m == 0 || m == n)    return 1;
    return 1LL*F[n] * inv_F[n - m] % mod * inv_F[m] % mod;
}
//end

int fib[N];
void init(){
    Mobius(N);
    init_C_small(2*N);
    fib[1]=fib[2]=1;
    for(int i=3;i<N;i++)fib[i]=(1LL*fib[i-1]+1LL*fib[i-2])%inv_mod;
}

int main(){
    init();
    int T;
    scanf("%d",&T);
    while(T--){
        int n,k;
        scanf("%d %d",&n,&k);
        LL ans=0;
        for(int i=1;i<=n;++i){
            if(n%i)continue;
            LL tmp=0;
            for(int j=i;j<=n;j+=i){
                if(n%j)continue;
                tmp=(tmp+1LL*mu[j/i]*C(n/j+k-1,k-1)%mod+mod)%mod;
            }
            ans=(ans+tmp*(pow_mod(2LL,fib[i],mod)-1+mod)%mod+mod)%mod;
        }
        LL ans2=C(n+k-1,k-1);
        //printf("ans1:%lld ans2:%lld\n",ans,ans2);
        ans=ans*pow_mod(ans2,mod-2,mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}

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