UVA11762 Race to 1 得到1

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这是一道挺不错的期望题。

令小于等于\(i\)中素数的个数为\(n\)，其中能整除\(i\)的有\(k\)个，那么可以列出一个期望方程：\(dp[i]=\frac{1}{n}\sum_{j=1}^{k}dp[\frac{i}{p_j}]+\frac{n-k}{n}dp[i]+1 \ (p_j|i)\).

化简得：$ dp[i] = \frac{1}{k}(\sum dp[\frac{i}{p_j}] + n)$.

于是就可以用记忆化搜索解决这道题了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e6 + 5;
In ll read()
{
	ll ans = 0;
	char ch = getchar(), las = ' ';
	while(!isdigit(ch)) las = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(las == '-') ans = -ans;
	return ans;
}
In void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
#endif
}

int n;

int v[maxn], sum[maxn], prm[maxn], pcnt = 0;
In void init()
{
	for(int i = 2; i < maxn; ++i)
		if(!v[i])
		{
			prm[++pcnt] = i;
			if(i < 1000) for(int j = i * i; j < maxn; j += i) v[j] = i;
		} 
	for(int i = 2; i < maxn; ++i) sum[i] = sum[i - 1] + (!v[i]);
}

db dp[maxn];
In db DP(int n)
{
	if(n == 1 || dp[n]) return dp[n];
	int m = n, cnt = 0;
	for(int i = 1; i <= pcnt && prm[i] <= m; ++i)
		if(m % prm[i] == 0)
		{	
			++cnt; 
			dp[n] += DP(n / prm[i]);
			while(m % prm[i] == 0) m /= prm[i];
		}
	return dp[n] = (dp[n] + sum[n]) / cnt;
}

int main()
{
//	MYFILE();
	init(); 
	int T = read();
	for(int id = 1; id <= T; ++id)
	{
		n = read();
		printf("Case %d: %.8lf\n", id, DP(n));
	}
	return 0;
}