UVa 11762 Race to 1 (数学期望 + 记忆化搜索)

时间：2022-09-11 04:48:47

题意：给定一个整数 n ，然后你要把它变成 1，变换操作就是随机从小于等于 n 的素数中选一个p，如果这个数是 n 的约数，那么就可以变成 n/p，否则还是本身，问你把它变成 1 的数学期望是多少。

析：一个很明显的期望DP，dp[i] 表示把 i 变成 1 的期望是多少，枚举每一种操作，列出表达式，dp[i] = ∑dp[i/x]/q + p/q*dp[i] + 1，其中 x 表示枚举的素数，然后 p 表示不是 i 的约数个数，q 是小于等于 n 的素数个数，然后变形，可以得到 dp[i] = (∑dp[i/x] + q) / (q-p)，可以用记忆化搜索来做。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

//#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.in", "r", stdin)

#define freopenw freopen("out.out", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1000000 + 10;

const int maxm = 100 + 2;

const LL mod = 100000000;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

double dp[maxn];

bool vis[maxn];

int prime[maxn], cnt;

double dfs(int x){

  if(x == 1)  return 0.;

  double &ans = dp[x];

  if(ans > 0.)  return ans;

  ans = 0.;

  int q = 0, p = 0;

  for(int i = 0; i < cnt && prime[i] <= x; ++i, ++q)

    if(x % prime[i] == 0)  dp[x] += dfs(x / prime[i]);

    else ++p;

  ans += q;

  ans /= q - p;

  return ans;

}

int main(){

  for(int i = 2; i < maxn; ++i)  if(!vis[i]){

    prime[cnt++] = i;

    if(i > 1000)  continue;

    for(int j = i*i; j < maxn; j += i)  vis[j] = 1;

  }

  int T;  cin >> T;  ms(dp, 0);

  for(int kase = 1; kase <= T; ++kase){

    scanf("%d", &n);

    printf("Case %d: %.6f\n", kase, dfs(n));

  }

  return 0;

}

　　

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