CF 441E

Description

一共執行\(k\)次，每次有\(p\%\)把\(x * 2\)，有\((100 - p)\%\)把\(x + 1\)。問二進制下\(x\)末尾期望\(0\)的個數。

Solution

設\(f[i][j]\)為執行第\(i\)次後\(x + j\)末尾期望\(0\)的個數

加一：$f[i + 1][j - 1] = f[i + 1][j - 1] + (100 - p)% * f[i][j]; $

乘二：\(f[i + 1][j * 2] = f[i + 1][j * 2] + p\% * (f[i][j] + 1);\)

#include<bits/stdc++.h>
using namespace std;
int x, k;
double p, p1;
double f[300][300];
int main() {
    scanf("%d%d%lf", &x, &k, &p);
    p /= 100, p1 = 1.0 - p;
    for (int i = 0; i <= k; i ++) 
        for (int j = x + i; j % 2 == 0; j /= 2)
            f[0][i] ++;
    for (int i = 0; i < k; i ++)
        for (int j = 0; j <= k; j ++) {
            if (j)
                f[i + 1][j - 1] += p1 * f[i][j]; // + 1
            if (j * 2 <= k)
                f[i + 1][j * 2] += p * (f[i][j] + 1); // * 2 
        }
    printf("%.10f\n", f[k][0]);
    return 0;
}
 

CF 441E Valera and Number