Valera is a coder. Recently he wrote a funny program. The pseudo code for this program is given below:

//input: integers x, k, p
a = x;
for(step = 1; step <= k; step = step + 1){
    rnd = [random integer from 1 to 100];
    if(rnd <= p)
        a = a * 2;
    else
        a = a + 1;
}

s = 0;

while(remainder after dividing a by 2 equals 0){
 
    a = a / 2;
    s = s + 1;
}

Now Valera wonders: given the values x, k and p, what is the expected value of the resulting number s?

The first line of the input contains three integers x, k, p (1 ≤ x ≤ 10⁹; 1 ≤ k ≤ 200; 0 ≤ p

Print the required expected value. Your answer will be considered correct if the absolute or relative error doesn‘t exceed 10^- 6.

1 1 50

1.0000000000000

5 3 0

3.0000000000000

5 3 25

1.9218750000000

題意：給定X，現在進行K輪操作，每一輪有P%的概率加倍，(100-P)%的概率加一，問K輪之後的X的因子2的次數。

思路：首先一個數X中2的因子個數=轉化為二進制後末尾0的個數=__builtin_ctz（X）；題解給的四維DP比較麻煩，這裏有一種比較難以想到，但是不難理解的二維DP。

用dp[i][j]，表示X+j進行i輪操作後的因子2的冪。 那麽答案就是dp[K][0]；

那麽轉移就是:

dp[i][j]+=(dp[i-1][j+1])*(1.0-P); 即第一次操作為+1；
dp[i][j<<1]+=(dp[i-1][j]+1)*P; 即第一次操作為*2；

雖然乘法的話第二維會加倍增長，但加法的話第二維會減小1，所以最小影響到dp[K][0]的最大第二維就是K，而不用維護所有的dp[][j]，即j<=K；

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std; //*2 P
double dp[210][410],P;
int main()
{
    int X,K;
    scanf("%d%d%lf",&X,&K,&P); P/=100.0;
    rep(i,0,K) dp[0][i]=__builtin_ctz(X+i);
    rep(i,1,K){
        rep(j,0,K){
            dp[i][j]+=(dp[i-1][j+1])*(1.0-P);
            dp[i][j<<1]+=(dp[i-1][j]+1)*P;
        }
    }
    printf("%.10lf\n",dp[K][0]);
    return 0;
}

CodeForces - 441E:Valera and Number (DP&數學期望&二進制)