HDU-6363：bookshelf（数论+容斥）

bookshelf
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)

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 $cnt_i(0≤cnt_i≤N)$ 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 $stable_i=f[cnt_i]$.

4. Define the beauty value of i-th layers $beauty_i=2^{stable_i}−1$.

5. Define the whole beauty value of ZYG $score=gcd(beauty_1,beauty_2,...,beauty_k)(Note: gcd(0,x)=x)$.

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

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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 $n,k(0<n,k≤10^6)$.

Output
For each test case, output the answer as a value of a rational number modulo $10^9+7$.

Formally, it is guaranteed that under given constraints the probability is always a rational number $\frac pq$ (p and q are integer and coprime, q is positive), such that q is not divisible by $10^9+7$. Output such integer a between 0 and $10^9+6$ that $p−aq$ is divisible by $10^9+7$.

Sample Input
1
6 8

Sample Output
797202805

思路：首先要知道2个公式：
$1.gcd(2^a-1,2^b-1)=2^{gcd(a,b)}-1$
$2.gcd(f[a],f[b])=f[gcd(a,b)]$
其中$f$数组代表斐波那契数列

那么所有的贡献是
$ans=\sum_{gcd=1}^{k|n} (2^{f[gcd]}-1)*(n本书分到k个书架且最大公约数为gcd的方案数)$。

其中$f[gcd]$可能很大，所以$2^{f[gcd]}-1$需要用欧拉降幂公式降幂。

因为是求期望，所以$ans=\frac {sum}{总方案数}$ 。

那么如何求n本书分到k个书架且大公约数为gcd的方案数呢？

其中n本书分到k个书架的总方案数为$C_{n+k-1}^{k-1}$。

那么n本书分到k个书架且$gcd$是$g$的倍数的方案数为$C_{\frac ng+k-1}^{k-1}$。

要求出$gcd$只为$g$的方案数就要用容斥了。

#include<bits/stdc++.h>
using namespace std;
const int MAX=2e6+10;
const int MOD=1e9+7;
typedef long long ll;
int f[MAX],v[MAX];
int isp[MAX],pr[MAX],all;
ll fac[MAX],inv[MAX];
ll POW(ll x,ll n,ll mod)
{
    ll res=1;
    while(n)
    {
        if(n&1)res=res*x%mod;
        x=x*x%mod;
        n/=2;
    }
    return res;
}
ll c(ll x,ll y){return fac[x]*inv[x-y]%MOD*inv[y]%MOD;}
ll cal(ll x){return (POW(2,f[x]+v[x]*(MOD-1),MOD)-1+MOD)%MOD;}//降幂
void init()            //预处理
{
    fac[0]=inv[0]=inv[1]=1;
    for(int i=1;i<=2e6;i++)fac[i]=fac[i-1]*i%MOD;
    for(int i=2;i<=2e6;i++)inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
    for(int i=2;i<=2e6;i++)inv[i]=inv[i]*inv[i-1]%MOD;
    f[0]=0,f[1]=f[2]=1;
    for(int i=3;i<=2e6;i++)
    {
        f[i]=f[i-1]+f[i-2];
        v[i]|=v[i-1];
        if(f[i]>=MOD-1)v[i]=1;
        f[i]%=MOD-1;
    }
    all = 0;
    memset(isp,0,sizeof isp);
    isp[1]=1;
    for(int i=2;i<MAX;i++)
    {
        if(!isp[i])pr[all++] = i;
        for(int j=0;j<all;j++)
        {
            long long t = 1LL*pr[j]*i ;
            if(t<MAX)
            {
                isp[t] = 1;
                if(i%pr[j]==0)break;
            }
            else break;
        }
    }
}
vector<int>p;
void Div(ll x)//分解成素因子
{
    for(ll i=2;i*i<=x;i++)
    {
        if(x%i==0)p.push_back(i);
        while(x%i==0)x/=i;
    }
    if(x>1)p.push_back(x);
}
ll solve(ll n,ll k)//容斥计算
{
    p.clear();
    Div(n);
    ll tot=c(n+k-1,k-1);
    for(int i=(1<<p.size())-1;i>=1;i--)
    {
        int tmp=1;
        for(int j=0;j<p.size();j++)if(i&(1<<j))tmp*=p[j];
        if(__builtin_popcount(i)%2)(tot-=c(n/tmp+k-1,k-1))%=MOD;
        else (tot+=c(n/tmp+k-1,k-1))%=MOD;
        (tot+=MOD)%=MOD;
    }
    return tot;
}
int main()
{
    init();
    int T;
    cin>>T;
    while(T--)
    {
        ll n,k,ans=0;
        scanf("%lld%lld",&n,&k);
        for(int i=1;i*i<=n;i++)//枚举公约数
        {
            if(n%i)continue;
            (ans+=solve(n/i,k)*cal(i)%MOD)%=MOD;
            if(n!=i*i)(ans+=solve(i,k)*cal(n/i)%MOD)%=MOD;
            (ans+=MOD)%=MOD;
        }
        ans=ans*POW(c(n+k-1,k-1),MOD-2,MOD)%MOD;
        printf("%lld\n",ans);
    }
    return 0;
}