BNU29140 Taikotaiko(概率)

BNU29140——Taiko taiko——————【概率题、规律题】

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

思路：应小妹的要求 讲清楚不装b
1 1*p+0*0.1(Y N)
2 3*p*p+1*p(1-p)+1*(1-p)*p+0(YY YN NY NN)==  P^2+2*p
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
double dp[1005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        double p;
        scanf("%d%lf",&n,&p);
//        printf("%lf%d")
        memset(dp,0,sizeof(dp));

        dp[1]=1.0*p;

        for(int i=2;i<=n;i++)
        dp[i]=(dp[i-1]+i)*p;

        printf("%.6lf\n",dp[n]);
    }
}