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*pn-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;
}