CF1097D Makoto and a Blackboard

题目链接

题意分析

首先我们令答案为\(dp[n][k]\)

经过观察可以发现答案是存在积性的

\[dp[n][k]* dp[m][k]=dp[n* m][k](gcd(n,m)==1) \]

那么为根据质数的唯一分解定理

\(x=p_1^{a_1}p_2^{a_2}......p_n^{a_n}\)

然后我们分别计算出\(p_x^{a_x}\)对应的答案 然后合并即可

我们令\(dp[i][j]\)表示当前某一个质数\(i\)次方分解\(j\)次的情况下的期望

\[dp[i][j]=\frac{1}{i+1}\sum_{k=0}^{i}dp[k][j-1] \]

直接暴搜+记忆化就可以了

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<string>
#include<queue>
#include<map>
#include<stack>
#include<list>
#include<set>
#include<deque>
#include<vector>
#include<ctime>
#define ll long long
#define inf 0x7fffffff
#define N 70
#define IL inline
#define M 10010
#define D double
#define mod 1000000007
#define R register
using namespace std;
template<typename T>IL void read(T &_)
{
    T __=0,___=1;char ____=getchar();
    while(!isdigit(____)) {if(____=='-') ___=0;____=getchar();}
    while(isdigit(____)) {__=(__<<1)+(__<<3)+____-'0';____=getchar();}
    _=___ ? __:-__;
}
/*-------------OI使我快乐-------------*/
ll n,k;
ll dp[N][M];
IL ll qpow(ll x,ll y)
{ll res=1;for(;y;y>>=1,x=x*x%mod) if(y&1) res=res*x%mod;return res;}
IL ll dfs(ll x,ll now,ll res)
{
    if(dp[now][res]!=-1) return dp[now][res];
    if(now==0) return dp[now][res]=1;
    if(res==0) return dp[now][res]=qpow(x,now);
    ll cdy=0;
    for(R ll i=0;i<=now;++i)
    cdy=(cdy+dfs(x,i,res-1))%mod;
    return dp[now][res]=cdy*qpow(now+1,mod-2)%mod;
}
IL ll solve(ll x)
{
    ll res=1;
    for(R ll i=2;i*i<=x;++i)
    {
        if(x%i==0)
        {
            ll cnt=0;
            while(x%i==0) x/=i,++cnt;
            memset(dp,-1,sizeof dp);
            res=res*dfs(i,cnt,k)%mod;
        }
    }
    memset(dp,-1,sizeof dp);
    if(x>1) res=res*dfs(x,1,k)%mod;
    return res;
}
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
    read(n);read(k);
    printf("%lld\n",solve(n));
//	fclose(stdin);
//	fclose(stdout);
    return 0;
}

HEOI 2019 RP++