UVa 11762 Race to 1 / 概率DP

公式f(x) = 1+f(x)*(1-g(x)/p(x))+ f(x/y)/p(x)

p(x)是不超过x的素数的个数

g(x)是p(x)个素数中是x因子的个数

然后移项（和我上次做的那一题差不多）

f(x)=(f(x/y)+p(x))/g(x)

y是x的因子 要枚举y

#include <cstdio>
#include <cstring>

const int maxn = 1000010;
int vis[maxn];
int prime[maxn];
int prime_cnt;
double f[maxn];
void sieve()
{
	for(int i = 2; i <= 1000; i++) 
		if(!vis[i])
			for(int j = i*i; j <= 1000000; j += i)
				vis[j] = 1;
}
void get_primes()
{
	sieve();
	prime_cnt = 0;
	for(int i = 2; i <= 1000000; i++)
		if(!vis[i])
			prime[prime_cnt++] = i;
}

double dp(int x)
{
	if(x == 1)
		return 0.0;
	if(vis[x])
		return f[x];
	vis[x] = 1;
	double& ans = f[x];
	ans = 0;
	int g = 0, p = 0;
	for(int i = 0; i < prime_cnt && prime[i] <= x; i++)
	{
		p++;
		if(x % prime[i] == 0)
		{
			g++;
			ans += dp(x / prime[i]);
		}
	}
	ans = (ans + p) / g;
	return ans;
}

int main()
{
	get_primes();
	memset(vis, 0, sizeof(vis));
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int n;
		scanf("%d", &n);
		printf("Case %d: ", cas++);
		printf("%.10f\n", dp(n));
	}
	return 0;
}

UVa 11762 Race to 1 / 概率DP

原文：http://blog.csdn.net/u011686226/article/details/18939643