Makoto and a Blackboard CodeForces - 1097D (积性函数dp)

大意: 初始一个数字$n$, 每次操作随机变为$n$的一个因子, 求$k$次操作后的期望值.

设$n$经过$k$次操作后期望为$f_k(n)$.

就有$f_0(n)=n$, $f_k(n)=\frac{\sum\limits_{d|n}{f_{k-1}(d)}}{\sigma_0(n)}, k>0$.

显然$f_k(n)$为积性函数, $dp$算出每个素因子的贡献即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e4+10;
int dp[N][70], sum[N][70], mi[N];

int DP(int p, int k, int r) {
	memset(dp,0,sizeof dp);
	memset(sum,0,sizeof sum);
	sum[0][0] = dp[0][0] = 1;
	REP(i,1,k) sum[0][i]=(sum[0][i-1]+(dp[0][i]=(ll)dp[0][i-1]*p%P))%P;
	REP(i,1,r) {
		sum[i][0] = dp[i][0] = 1;
		REP(j,1,k) sum[i][j]=(sum[i][j-1]+(dp[i][j]=sum[i-1][j]*inv(j+1)%P))%P;
	}
	return dp[r][k];
}

int main() {
	int k;
	ll n;
	scanf("%lld%d", &n, &k);
	int mx = sqrt(n+0.5), ans = 1;
	REP(i,2,mx) if (n%i==0) {
		int x = 0;
		while (n%i==0) n/=i, ++x;
		ans = (ll)ans*DP(i,x,k)%P;
	}
	if (n>1) ans = (ll)ans*DP(n%P,1,k)%P;
	if (ans<0) ans+=P;
	printf("%d\n", ans);
}