HDU-6363 bookshelf 莫比乌斯反演
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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 :
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Assume that the i-th layer has cnti(0≤cnti≤N) books finally.
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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]).
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Define the stable value of i-th layers stablei=f[cnti].
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Define the beauty value of i-th layers beautyi=2stablei−1.
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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 ≤ 1 0 6 ) . (0<n,k\le10^{6}). (0<n,k≤106).
Output
For each test case, output the answer as a value of a rational number modulo 1 0 9 + 7. 10^{9}+7. 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 1 0 9 + 6. 10^{9}+6. 109+6. that p−aq is divisible by 1 0 9 + 7. 10^{9}+7. 109+7.
Sample Input
1 6 8
Sample Output
797202805
**题意：**把N本书放到K层的书架上，每一层的美丽值为 b i = 2 f i b [ c n t ] − 1 b_{i}=2^{fib[cnt]}-1 bi=2fib[cnt]−1，其中cnt是这一层书的数量，fib[x]为斐波那契数列，整个书架的美丽值为 g c d ( b 1 , b 2 , . . . , b k ) gcd(b_{1},b_{2},...,b_{k}) gcd(b1,b2,...,bk)，问整个书架的美丽值的期望
**题解：**考试的时候因为本人实在是太菜了。对这个题目完全没有一点思路，考完试以后听了dls直播算是对这个题目有了一点思路代码很短，但是从学习各种姿势到写完用了一天，感觉一下子掌握了好多神奇的姿势，开心。
经典定理： g c d ( x a − 1 , x b − 1 ) = x g c d ( a , b ) − 1 \large gcd(x^{a}-1,x^{b}-1)=x^{gcd(a,b)}-1 gcd(xa−1,xb−1)=xgcd(a,b)−1
g c d ( f i b [ a ] , f i b [ b ] ) = f i b [ g c d ( a , b ) ] \large gcd(fib[a],fib[b])=fib[gcd(a,b)] gcd(fib[a],fib[b])=fib[gcd(a,b)]
那么对于这个书架的美丽值可以做如下变形：
\begin{align*} \large gcd(b_{1},b_{2},…,b_{k})
&\Large= gcd(2{fib[x_{1}]},2{fib[x_{2}]},…,2^{fib[x_{k}]})\
&\Large= 2{gcd(fib[x_{1}]},2{fib[x_{2}]},…,2^{fib[x_{k}])}\
&\Large=2^{fib[gcd(x_{1},x_{2},……,x_{1k})]}
\end{align*}
于是我们设 g = g c d ( x 1 , x 2 , … … , x k ) \large g=gcd(x_{1},x_{2},……,x_{k}) g=gcd(x1,x2,……,xk)
那么整个书架的美丽值变为 2 f i b [ g ] … … ① \large 2^{fib[g]}……① 2fib[g]……①
又因为
\begin{align*}\large N&=x_{1}+x_{2}……+x_{k}\
\large&=ga_{1}+ga_{2}……+ga_{k}\
\large&=g(a_{1}+a_{2}……+a_{k})……②\
\end{align*}
显然g为N的约数而根据①式可得书架美丽值得期望仅与g有关
于是我们枚举N的每一个约数g并统计约数为g的方案数，可以由简单的期望公式就算出答案。
对于g的方案数我们可以对上述②继续变形
\begin{align*}\large N&=g*(a_{1}+a_{2}……+a_{k})\
\frac{N}{g}&=a_{1}+a_{2}……+a_{k}……③
\end{align*}
所以方案数就等于③式非负整数解的组数。根据组合数知识我们可以很容易得到③式非负整数解的组数为
C N g + K − 1 K − 1 \Large C^{K-1}_{\frac{N}{g}+K-1} CgN+K−1K−1
需要注意的是若xg（x为大于1的整数）也为N的约数那么③式存在这样的解
\begin{align}
\frac{N}{g}&=xa_{1}+xa_{2}……+xa_{k}\
\frac{N}{g}&=x(a_{1}+a_{2}……+a_{k})\
\frac{N}{xg}&=a_{1}+a_{2}……+a_{k}\
\end{align}
所以N的x*g这个因子被重复计算了多次。简单的容斥原理
若我们设 f [ g ] f[g] f[g]为严格约数g的方案数， F [ g ] F[g] F[g]为非严格约数g的方案数。那么我们可以很容易得到以下式子
f [ g ] = F [ g ] − F [ 2 g ] − F [ 3 g ] − F [ 5 g ] + F [ 6 g ] … … F [ N ] f[g]=F[g]-F[2g]-F[3g]-F[5g]+F[6g]……F[N] f[g]=F[g]−F[2g]−F[3g]−F[5g]+F[6g]……F[N]
那么我们用莫比乌斯函数总结以上表达式可以得到
f [ g ] = ∑ g ∣ d & d ∣ N μ ( d g ) F [ d ] \large f[g] = \sum_{g|d\&d|N}\mu{(\frac{d}{g})}F[d] f[g]=g∣d&d∣N∑μ(gd)F[d]
而
F [ g ] = C N g + K − 1 K − 1 \Large F[g]= C^{K-1}_{\frac{N}{g}+K-1} F[g]=CgN+K−1K−1
这样我们就得到了美丽值为 2 f i b [ g ] 2^{fib[g]} 2fib[g]的方案数为$ f[i]$
理论上我们只要将两者相乘，然后除于总共方案数就可以了。但是 f i b [ g ] fib[g] fib[g]是一个指数级增长的函数式，所以我们需要利用欧拉函数对这个式子进行优化
欧拉函数
若 （ a , m ） = 1 则 a ϕ ( m ) ≡ 1 ( m ) 若（a,m）=1则a^{\phi(m)} \equiv1(m) 若（a,m）=1则aϕ(m)≡1(m)
所以在本题中 m = 1 0 9 + 7 m=10^{9}+7 m=109+7
2 m − 1 ≡ 1 ( m o d m ) 2^{m-1} \equiv1(mod\ m) 2m−1≡1(mod m)
那么 2 f i b [ i ] ≡ 2 f i b [ i ] % ( m − 1 ) ( m o d m ) \large2^{fib[i]} \equiv2^{fib[i]\%(m-1)} (mod\ m) 2fib[i]≡2fib[i]%(m−1)(mod m)
这样我们就可以很容易的通过莫比乌斯函数，算出严格约数g的方案数。
我已经对本题做了简单的数学推导。我们只要把上文中的内容简单预处理然后实现一下就好了。唯一需要注意的是本题空间有一点卡。所以有几个函数要用int而不能一股脑都用longlong
#include
const long long MAXN = 1000000,MOD = 1000000007;
using namespace std;
long long T,ans,N,K,cnt;
int mu[MAXN+5],prime[MAXN+5];
long long fnv[2*MAXN+5],fac[2*MAXN+5],inv[MAXN+5],f[MAXN+5],fib[MAXN+5];
bool vis[MAXN+5];
vectord;
inline long long powmod(long long a,long long b){
long long res=1;
a%=MOD;
for(;b;b>>=1){
if(b&1)res=res*a%MOD;
a=a*a%MOD;
}
return res;
}
inline void pre(){
fib[0] = 0;fib[1] = 1;inv[1] = 1;
fac[0] = fnv[0] = 1;
for(int i = 2;i <= MAXN;i++) fib[i] = (fib[i-1] + fib[i-2]) % (MOD - 1);
for(int i = 2;i <= MAXN;i++) inv[i] = (MOD-MOD/i)*inv[MOD%i]%MOD;
for(int i = 1;i<=2*MAXN;i++) {
fac[i] = fac[i-1] * i % MOD;
fnv[i] = fnv[i-1] * inv[i] % MOD;
}
mu[1] = 1;cnt = 0;
for(int i = 2;i <= MAXN;i++){
if(!vis[i]){
prime[cnt++] = i;
mu[i] = -1;
}
for(int j = 0;j < cnt&&i*prime[j] <= MAXN;j++){
vis[i*prime[j]] = true;
if(i%prime[j]) mu[i*prime[j]] = -mu[i];
else {
mu[i*prime[j]] = 0;
break;
}
}
}
}
inline long long comb(long long a,long long b){
return ((fac[a]*fnv[b])%MOD*fnv[a-b])%MOD;
}
int main()
{
pre();
scanf("%lld",&T);
while(T--){
scanf("%lld%lld",&N,&K);
d.clear();memset(f,0,sizeof(f));
for(int i = 1;i <= N;i++) if(N%i == 0) d.push_back(i);
for(int i = 0;i < d.size();i++)
for(int j = i;j < d.size();j++)
if(d[j]%d[i] == 0){
f[i] = ((f[i] + mu[d[j]/d[i]]*comb(N/d[j]+K-1,N/d[j]))%MOD + MOD)%MOD;
}
ans = 0;
for(int i = 0;i < d.size();i++) ans=(ans+f[i]*(powmod(2,fib[d[i]])-1))%MOD;
ans = ans * powmod(comb(N+K-1,K-1),MOD-2) % MOD;
if(ans < 0) ans += MOD;
printf("%lld\n",ans);
}
return 0;
}
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