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 N book .
The ZYG has K K layers (count from 1 1 to K K ) and there is no limit on the capacity of each layer !
Now Patrick want to put all N N books on ZYG :
Assume that the i-th layer has cnti(0≤cnti≤N) c n t i ( 0 ≤ c n t i ≤ N ) books finally.
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]). ( f [ 0 ] = 0 , f [ 1 ] = 1 , f [ 2 ] = 1 , f [ i ] = f [ i − 2 ] + f [ i − 1 ] ) .
Define the stable value of i-th layers stablei=f[cnti] s t a b l e i = f [ c n t i ] .
Define the beauty value of i-th layers beautyi=2stablei−1 b e a u t y i = 2 s t a b l e i − 1 .
Define the whole beauty value of ZYG score=gcd(beauty1,beauty2,...,beautyk)(Note:gcd(0,x)=x) s c o r e = g c d ( b e a u t y 1 , b e a u t y 2 , . . . , b e a u t y k ) ( N o t e : g c d ( 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) n , k ( 0 < n , k ≤ 10 6 ) .
Output
For each test case, output the answer as a value of a rational number modulo 109+7 10 9 + 7 .
Formally, it is guaranteed that under given constraints the probability is always a rational number pq p q (p and q are integer and coprime, q is positive), such that q is not divisible by 109+7 10 9 + 7 . Output such integer a between 0 and 109+6 10 9 + 6 that p−aq p − a q is divisible by 109+7 10 9 + 7 .
Sample Input
1
6 8
Sample Output
797202805
思路：首先要知道2个公式：
1.gcd(2a−1,2b−1)=2gcd(a,b)−1 1. g c d ( 2 a − 1 , 2 b − 1 ) = 2 g c d ( a , b ) − 1
2.gcd(f[a],f[b])=f[gcd(a,b)] 2. g c d ( f [ a ] , f [ b ] ) = f [ g c d ( a , b ) ]
其中 f f 数组代表斐波那契数列
那么所有的贡献是
ans=∑gcd|ngcd=1(2f[gcd]−1)∗(n本书分到k个书架且最大公约数为gcd的方案数) a n s = ∑ g c d = 1 g c d | n ( 2 f [ g c d ] − 1 ) ∗ ( n 本 书 分 到 k 个 书 架 且 最 大 公 约 数 为 g c d 的 方 案 数 ) 。
其中 f[gcd] f [ g c d ] 可能很大，所以 2f[gcd]−1 2 f [ g c d ] − 1 需要用欧拉降幂公式降幂。
因为是求期望，所以 ans=ans总方案数 a n s = a n s 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 Ck−1n+k−1 C n + k − 1 k − 1 。
那么n本书分到k个书架且 gcd g c d 是 g g 的倍数的方案数为 Ck−1ng+k−1 C n g + k − 1 k − 1 。
要求出 gcd g c d 只为 g g 的方案数就要用容斥了。
#includeif(!isp[i])pr[all++] = i;
for(int j=0;j1LL*pr[j]*i ;
if(t1;
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<1;i>=1;i--)
{
int tmp=1;
for(int j=0;jif(i&(1<*=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;
}