[Codeforces441E]Valera and Number

Problem

给定一个数x，有p%的概率乘2，有1-p%的概率加1，问操作k次，其二进制数下末尾零的个数的期望。

Solution

每次操作只会影响到最后的8位
我们用dp[i][j]表示i个操作后，后面的操作还需要加j
对于+1的操作：dp[i][j-1]+=dp[i-1][j](1-p)
对于2的操作：dp[i][j2]+=(dp[i-1][j]+1)p

Notice

需要预处理

Code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9;
const double eps = 1e-6, phi = acos(-1.0);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
double p;
int x, k;
double f[205][205];
int sqz()
{
	int x = read(), k = read();
	scanf("%lf", &p), p /= 100;
	rep(i, 0, k)
	{
		int t = x + i;
		while (1)
		{
			if (t & 1) break;
			f[0][i]++;
			t /= 2;
		}
	}
	rep(i, 1, k)
	{
		memset(f[i], 0, sizeof f[i]);
		rep(j, 0, k)
		{
			if (j) f[i][j - 1] += f[i - 1][j] * (1 - p);
			if (j * 2 <= k) f[i][j * 2] += (f[i - 1][j] + 1) * p;
		}
	}
	printf("%.10lf\n", f[k][0]);
}