LightOJ 1038 Race to 1 Again

期望，$dp$。

设$ans[n]$为答案，那么$ans[n]=(ans[a]+ans[b]+ans[c]+.....+ans[n])/cnt+1$，其中$a$,$b$,$c$......$n$均为$n$的因子。

化简得到$ans[n]=(ans[a]+ans[b]+ans[c]+......+cnt)/(cnt-1)$。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
}

int T,n;
double ans[100010];
vector<int>v[100010];

int main()
{
    for(int i=1;i<=100000;i++)
    {
        for(int j=1;;j++)
        {
            if(i*j>100000) break;
            v[i*j].push_back(i);
        }
    }

    for(int i=2;i<=100000;i++)
    {
        for(int j=0;j<v[i].size();j++)
            ans[i]=ans[i]+ans[i/v[i][j]];
        ans[i]=ans[i]+v[i].size();
        ans[i]=ans[i]/(v[i].size()-1);
    }

    scanf("%d",&T); int cas=1;
    while(T--)
    {
        scanf("%d",&n);
        printf("Case %d: %lf
",cas++,ans[n]);
    }
    return 0;
}