uva11762 Race to 1

Dilu have learned a new thing about integers, which is - any positive
integer greater than 1 can be divided by at least one prime number
less than or equal to that number. So, he is now playing with this
property. He selects a number N . And he calls this D . In each turn
he randomly chooses a prime number less than or equal to D . If D is
divisible by the prime number then he divides D by the prime number to
obtain new D . Otherwise he keeps the old D . He repeats this
procedure until D becomes 1. What is the expected number of moves
required for N to become 1. [We say that an integer is said to be
prime if its divisible by exactly two different integers. So, 1 is not
a prime, by de nition. List of rst few primes are 2, 3, 5, 7, 11,
…] Input Input will start with an integer T ( T  1000), which
indicates the number of test cases. Each of the next T lines will
contain one integer N (1  N  1000000). Output For each test case
output a single line giving the case number followed by the expected
number of turn required. Errors up to 1e-6 will be accepted.

设dp(i)表示从i开始的期望步数，有n个不超过i的质数，其中有m个是i的因数，分别为a1..am，那么有

dp(i)=1+(1−mn)∗dp(i)+∑j=1mdp(iaj)∗1n

化简得

dp(i)=∑mj=1dp(iaj)+nm

记忆化搜索。

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define LL long longconst int maxx=1000000;int prm[1000010],tot,have[1000010];double dp[1000010];double dfs(int x){    if (x==1) return 0;    if (dp[x]) return dp[x];    int m=0,n=0,i;    for (i=1;i<=tot&&prm[i]<=x;i++)    {        n++;        if (x%prm[i]==0) dp[x]+=dfs(x/prm[i]),m++;    }    return dp[x]=(dp[x]+n)/m;}int main(){    int T,K,i,j,x;    for (i=2;i<=maxx;i++)    {        if (!have[i]) prm[++tot]=i;        for (j=1;j<=tot&&(LL)prm[j]*i<=maxx;j++)        {            have[prm[j]*i]=1;            if (i%prm[j]==0) break;        }    }    scanf("%d",&T);    for (K=1;K<=T;K++)    {        scanf("%d",&x);        printf("Case %d: %.10f\n",K,dfs(x));    }}