标签：概率dp
| Time Limit: 2 second(s) | Memory Limit: 32 MB |
Rimi learned a new thing about integers, which is - any positive integer greater than 1 can be divided by its divisors. So, he is now playing with this property. He selects a number N. And he calls thisD.
In each turn he randomly chooses a divisor of D (1 to D). Then he divides D by the number to obtain new D. He repeats this procedure until D becomes 1. What is the expected number of moves required for N to become 1.
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case begins with an integer N (1 ≤ N ≤ 105).
For each case of input you have to print the case number and the expected value. Errors less than 10-6 will be ignored.
Sample Input |
Output for Sample Input |
|
3 1 2 50 |
Case 1: 0 Case 2: 2.00 Case 3: 3.0333333333 |
设x有n个因子，dp[x] =(dp[i]+dp[j]+....+dp[k])*(1/n)+dp[n]*1/n+1; (i,j,k表示x的因子)
换一下就可以得到dp[x]的表达式了，
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
#define eps 1e-8
//typedef __int64 ll;
#define fre(i,a,b) for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define ssf(n) scanf("%s", n)
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define bug pf("Hi\n")
using namespace std;
#define INF 0x3f3f3f3f
#define N 100005
double dp[N];
int n;
void inint()
{
int i,j,cnt;
double temp;
dp[1]=0;
mem(dp,0);
fre(i,2,N)
{
cnt=0;
temp=0;
for(j=1;j*j<=i;j++)
if(i%j==0)
{
cnt++;
temp+=dp[j];
if(j*j!=i)
{
temp+=dp[i/j];
cnt++;
}
}
dp[i]=(temp+cnt)/(cnt-1);
}
}
int main()
{
int i,j,t,ca=0;
sf(t);
inint();
while(t--)
{
sf(n);
pf("Case %d: %.6lf\n",++ca,dp[n]);
}
return 0;
}
Light oj 1038 Race to 1 Again（概率dp）
标签：概率dp
原文地址：http://blog.csdn.net/u014737310/article/details/45342465
