BNU29140——Taiko taiko——————【概率题、规律题】

最新推荐文章于 2025-08-08 22:01:10 发布

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最新推荐文章于 2025-08-08 22:01:10 发布

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文章标签： java

原文链接：http://www.cnblogs.com/chengsheng/p/4397406.html

本文介绍了一种简化版的太鼓达人游戏积分计算方法，针对不同击打序列和击中概率，通过数学公式计算玩家在给定长度的序列中获得积分的期望值。

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Taiko taiko

Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld Java class name: Main

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拆拆超级喜欢太鼓达人（赛后大家可自行百度规则），玩久了也对积分规则产生了兴趣，理论上连击数越多，分数增加的越快，而且还配合着击打准确度有相应的计算规则，拆拆觉得这些规则太复杂了，于是把规则自行简化了下：

对于一段击打序列，我们假设Y为打中，N为未打中（没有良可之分了）

我们视连续的n次击中为n连击相应的分数为 1+2+3+。。。+n

例如序列YNNYYYNYN的总分数为1+1+2+3+1=8

当然击中是有概率的我们假定概率始终为P（0<=P<=1）拆拆的击中概率很高的恩恩=w=

于是现在拆拆想知道对于长度为L的序列击中概率为P时获得积分的期望是多少

Input

一个整数T（表示T组数据）

接下来的T组数据

接下来T行每行一个整数L 一个浮点数P

数据范围

1<=T<=1000

1<=L<=1000

0<=P<=1

Output

对于每组数据输出一行1个6位小数即题目描述的期望

Sample Input

2
2 0.9
3 0.5

Sample Output

2.610000
2.125000



解题思路：应该是有证明和推导的，但是好多人都是找规律找出来的，规律就是：1*pⁿ+2*p^n-1+...n*p。至于为啥这样，数学渣也表示很无语。

#include<bits/stdc++.h>
using namespace std;
int main(){

    int t;
    scanf("%d",&t);
    while(t--){

        int n;
        double p,ans=0,tmp;
        scanf("%d%lf",&n,&p);
        tmp=p;
        for(int i=n;i>0;i--){

            ans+=i*tmp;
            tmp*=p;
        }
        printf("%.6lf\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/chengsheng/p/4397406.html