【精度控制+log应用】Candy_控制log精度-优快云博客

解决一个有趣的概率问题：一个懒惰的小孩有两个装满糖果的盒子，每天随机选择一个盒子吃一颗糖，直到有一天发现其中一个盒子空了。通过编程计算另一个盒子中剩余糖果的期望数量。

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Candy

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3072 Accepted Submission(s): 1418
Special Judge

Problem Description

LazyChild is a lazy child who likes candy very much. Despite being very young, he has two large candy boxes, each contains n candies initially. Everyday he chooses one box and open it. He chooses the first box with probability p and the second box with probability (1 - p). For the chosen box, if there are still candies in it, he eats one of them; otherwise, he will be sad and then open the other box.
He has been eating one candy a day for several days. But one day, when opening a box, he finds no candy left. Before opening the other box, he wants to know the expected number of candies left in the other box. Can you help him?

Input

There are several test cases.
For each test case, there is a single line containing an integer n (1 ≤ n ≤ 2 × 105) and a real number p (0 ≤ p ≤ 1, with 6 digits after the decimal).
Input is terminated by EOF.

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is a real number indicating the desired answer.
Any answer with an absolute error less than or equal to 10-4 would be accepted.

Sample Input

10 0.400000 100 0.500000 124 0.432650 325 0.325100 532 0.487520 2276 0.720000

Sample Output

Case 1: 3.528175 Case 2: 10.326044 Case 3: 28.861945 Case 4: 167.965476 Case 5: 32.601816 Case 6: 1390.500000

#include<bits/stdc++.h>
using namespace std;
double c[400005];
void init() {
    memset(c,0,sizeof(c));
    for(int i=2; i<=400000; i++) {
        c[i]=c[i-1]+log(i);
    }
}
int n;
double deal(double p) {
    double ans=0;
    for(int i=0; i<n; i++) {
        ans+=(n-i)*exp((n+1)*log(p)+(i)*log(1-p)+(c[n+i]-c[i]-c[n]));
    }
    return ans;
}
int main() {
    init();
    int cas=1;
    while(cin>>n) {
        double p;
        cin>>p;
        double ans;
        if (p==0||p==1) ans=n;
        else ans=deal(p)+deal(1-p);
        cout<<"Case "<<cas<<": ";
        printf("%.6lf\n",ans);
        cas++;
    }

}