UVa 11762 - Race to 1 （概率 期望 DP 馬爾可夫過程）

UVA - 11762

Race to 1

Time Limit: 10000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

B

Race to 1 Input: Standard Input Output: Standard Output

 

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 newD. 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 definition. List of first 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.

 

Sample Input                             Output for Sample Input

3 1 3 13

Case 1: 0.0000000000 Case 2: 2.0000000000 Case 3: 6.0000000000

---

Problemsetter: Md. Arifuzzaman Arif

Special Thanks: Sohel Hafiz

Source

Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 2. Mathematics :: Probability :: Examples
Root :: Prominent Problemsetters :: Md. Arifuzzaman Arif

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題意：

給出一個整數n，每次可以在不超過n的素數中選擇一個p，如果p是n的約數，則把n變成n/p，否則n不變。問平均情況下，需要多少次隨機選擇才能把n變爲1。

隨機狀態轉移機

這樣的隨機過程稱爲馬爾可夫過程(markov process)

設f(x)表示x變成1的隨機選擇期望次數

根據全期望可得：

f(x) = f(x) * (1 - g(x)/p(x)) + ∑(f(x/i) * (1/p(x)) + 1

其中p(x)表示不超過x的素數數量，g(x)表示不超過x的素數並且是x約數的數量

i爲不超過x的素數並且是x約數

因爲等式兩邊都有f(x)，移項整理可得：

f(x) = ( p(x) + ∑f(x/i) ) / g(x) 

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const double PI = (4.0*atan(1.0));

const int maxn = 1000000 + 20;
const int maxp = 100000;
int vis[maxn];
int prime[maxp];
double dp[maxn];
int primNum;

void sieve(int n) {
    int m = (int) sqrt(n + 0.5);
    memset(vis, 0, sizeof(vis));
    for(int i=2; i<=m; i++) if(!vis[i]) {
        for(int j=i*i; j<=n; j+=i) vis[j] = 1;
    }
}

int getPrimes(int n) {
    sieve(n);
    int c = 0;
    for(int i=2; i<=n; i++) if(!vis[i])
        prime[c++] = i;
    return c;
}

double f(int n) {
    if(n == 1) return 0;
    if(vis[n]) return dp[n];
    vis[n] = 1;
    int p = 0;
    int g = 0;
    double ans = 0;
    for(int i=0; i<primNum && prime[i] <= n; i++) {
        p++;
        if(n % prime[i] == 0) {
            g++;
            ans += f(n/prime[i]);
        }
    }
    return dp[n] = (p + ans) / g;
}

int main() {
    int T;

    primNum = getPrimes(maxn-1);
    memset(dp, 0, sizeof(dp));
    memset(vis, 0, sizeof(vis));
    scanf("%d", &T);
    for(int kase=1; kase<=T; kase++) {
        int n;
        scanf("%d", &n);
        double ans = f(n);
        printf("Case %d: %.10lf\n", kase, ans);
    }

    return 0;
}