Light OJ 1038 Race to 1 Again (probabilistic DP)

1038-race to 1 Again

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Time Limit: 2 second (s)

Memory Limit: MB

Rimi learned a new thing about integers, which is-any positive integer greater than 1 can is divided by its div Isors. So, he's now playing with the 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's the expected number of moves required for N to become 1.

Input

Input starts with an integer T (≤10000), denoting the number of test cases.

Each case is begins with an integer N (1≤n≤105).

Output

For each case, the input you has to print the case number and the expected value. Errors less than 10-6 'll be ignored.

Sample Input	Output for Sample Input
3 1 2 50	Case 1:0 Case 2:2 Case 3:3.0333333333

Problem Setter:jane ALAM JAN

Set x has n factors, dp[x] = (Dp[i]+dp[j]+....+dp[k]) * (1/n) +dp[n]*1/n+1; (I,j,k = factor of X)

Change it to get dp[x] expression,

#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) MEMS ET ((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 100005double dp[n];int n;void inint () {int i,j,cnt;d   Ouble 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);p f ("Case%d:%.6lf\n", ++ca,dp[n]);} return 0;}

Light OJ 1038 Race to 1 Again (probabilistic DP)