D. Makoto and a Blackboard
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Makoto has a big blackboard with a positive integer nn written on it. He will perform the following action exactly kk times:
Suppose the number currently written on the blackboard is vv. He will randomly pick one of the divisors of vv (possibly 11 and vv) and replace vv with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 5858 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after kk steps.
It can be shown that this value can be represented as PQPQ where PP and QQ are coprime integers and Q≢0(mod109+7)Q≢0(mod109+7). Print the value of P⋅Q−1P⋅Q−1 modulo 109+7109+7.
Input
The only line of the input contains two integers nn and kk (1≤n≤10151≤n≤1015, 1≤k≤1041≤k≤104).
Output
Print a single integer — the expected value of the number on the blackboard after kk steps as P⋅Q−1(mod109+7)P⋅Q−1(mod109+7) for PP, QQ defined above.
Examples
input
Copy
6 1
output
Copy
3
input
Copy
6 2
output
Copy
875000008
input
Copy
60 5
output
Copy
237178099
Note
In the first example, after one step, the number written on the blackboard is 11, 22, 33 or 66 — each occurring with equal probability. Hence, the answer is 1+2+3+64=31+2+3+64=3.
In the second example, the answer is equal to 1⋅916+2⋅316+3⋅316+6⋅116=1581⋅916+2⋅316+3⋅316+6⋅116=158.
题意:一开始你手中有一个数n,每一轮你都会等概率的把这个数变成它的因子,k轮后求这个数大小的期望。
题解:首先这个期望是一个积性函数(???为啥,大佬说是就是吧),那么对于一个积性函数,他的期望可以变为他各个质因子期望的乘积,设dp[i][j]代表第i轮这个质因子的幂为j的期望次数,设上一轮的幂为k,那么这一轮的j都会由上一轮转换的概率为1/(k+1),处理期望dp即可
#include<bits/stdc++.h>
using namespace std;
#define Sheryang main
const int maxn=1e5+7;
typedef long long ll;
const int mod=1e9+7;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();if(c == '-')Nig = -1,c = getchar();while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();return Nig*x;}
#define read read()
/* keep hungry and keep calm! */
ll dp[(int)1e4+7][62],n,k,inv[62];
ll qpow(ll a,ll b){
ll ans=1;a%=mod;
while(b){
if(b&1) ans=(ans*a)%mod;
a=(a*a)%mod;b/=2;
}
return ans;
}
ll solve(ll p,int e){
dp[0][e]=1;
for(int i=0;i<e;i++) dp[0][i]=0;
for(ll i=1;i<=k;i++){
for(int j=0;j<=e;j++){
dp[i][j]=0;
for(int k=j;k<=e;k++){
dp[i][j]=(dp[i][j]+dp[i-1][k]*inv[k+1]%mod)%mod;
}
}
}
ll tmp=0,t=1;
for(int i=0;i<=e;i++){
tmp=(tmp+t*dp[k][i]%mod)%mod;
t=(t*p)%mod;
}
return tmp;
}
int Sheryang()
{
inv[1]=1;
for(int i=2;i<=60;i++){
inv[i]=qpow(i,mod-2);
}
n=read,k=read;
ll ans=1;
for(ll i=2;i*i<=n;i++){
int cot=0;
while(n%i==0) cot++,n/=i;
if(cot) ans=(ans*solve(i,cot)%mod)%mod;
}
if(n>1) ans=(ans*solve(n,1)%mod)%mod;
printf("%lld\n",ans);
return 0;
}