14030: 【原4030】cfcome

题目

题目描述

author: Kodak 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/4030 

Description

日天最近入坑了一个隔膜叫cfcome，因为一直打不过wtz还充钱开了svip

vip可以让日天知道自己每一枪开出去打中wtz的概率，svip在此基础上增加了伤害，不再每枪造成1点伤害，而是连续击中的第K枪造成2K-1点伤害，如果没击中则重新开始计数

然而日天还是只会见面把弹夹打空，他想知道有没有可能打赢wtz，所以需要计算自己对wtz的期望伤害

Input Format

第一行一个数N，N <= 1e6, 日天弹夹里剩余子弹数。

之后一行N个用空格分开的实数ai， (0<=ai<=1)， 每一枪打中wtz的几率

Output Format

一个数，日天对wtz的期望伤害，保留六位小数

Sample Input

3
0.5 0.5 0.5

Sample Output

2.750000

Pangbo13's solution

/*
一道数学题
*/
#include<iostream>
using namespace std;
int main(){
    double damage = 0;
    double last_hit_pro = 0,last_damage_avg = 0;
    double this_hit_pro = 0,this_damage_avg = 0;
    int n;
    scanf("%d",&n);
    for(int i = 0;i<n;i++){
        scanf("%lf",&this_hit_pro);
        this_damage_avg = (last_damage_avg+2 * last_hit_pro) * this_hit_pro + (1-last_hit_pro) * this_hit_pro;
        last_hit_pro = this_hit_pro;
        last_damage_avg = this_damage_avg;
        damage += this_damage_avg;
    }
    printf("%f",damage);
}

q4x3's solution

/**
 * dp
 * f[i]表示第i颗子弹对答案的贡献
 * 状态转移方程:
 * f[i] = ((1 - a[i - 1]) + a[i - 1] * (f[i - 1] / a[i - 1] + 2)) * a[i]
 * 化简
 * 注意输出格式
 **/
#include <iostream>
#include <iomanip>

using namespace std;

int N;
double a[100233], f[100233], ans;

int main() {
    cin >> N;
    for(int i = 1;i <= N;++ i)
        cin >> a[i];
    f[1] = a[1];
    ans = a[1];
    for(int i = 2;i <= N;++ i) {
        f[i] = (1 + f[i - 1] + a[i - 1]) * a[i];
        ans += f[i];
    }
    cout << fixed << setprecision(6) << ans << endl;
}

victrid's solution

#include <iomanip>
#include <iostream>
using namespace std;
int main() {
    int N;
    cin >> N;
    double* P = new double[N + 1];
    //single shot
    double* Eh = new double[N + 1];
    double* E  = new double[N + 1];
    for (int i = 1; i <= N; i++) {
        cin >> P[i];
    }
    P[0]  = 0;
    E[0]  = 0;
    Eh[0] = 0;
    for (int i = 1; i <= N; i++) {
        Eh[i] = (Eh[i - 1] + 1) * P[i - 1] + 1;
        E[i]  = E[i - 1] + Eh[i] * P[i];
    }
    cout << setiosflags(ios::fixed) << setprecision(6) << E[N];
    return 0;
}

zqy2018's solution

#include <bits/stdc++.h>
using namespace std;
int main(){
    int n;
    scanf("%d", &n);
    if (!n) {
        printf("%.6lf\n", 0.0);
        return 0;
    }
    double sum1 = 0, sum2 = 0, ans = 0, lst = 0;
    for (int i = 1; i <= n; ++i){
        double p;
        scanf("%lf", &p);
        sum1 += 2 * sum2, sum1 += (1 - lst), sum1 *= p;
        ans += sum1;
        sum2 += (1 - lst), sum2 *= p;
        lst = p;
    }
    printf("%.6lf\n", ans);
    return 0;
}