p(k)=(y^k) / (k!) * e^(-y)

k就是我们要求的大小。

Alien's Organ

Time Limit: 2 Seconds      Memory Limit: 65536 KB

There's an alien whose name is Marjar. It is an universal solder came from planet Highrich a long time ago.

Marjar is a strange alien. It needs to generate new organs(body parts) to fight. The generated organs will provide power to Marjar and then it will disappear. To fight for problem of moral integrity decay on our earth, it will randomly generate new fighting organs all the time, no matter day or night, no matter rain or shine. Averagely, it will generate λ new fighting organs every day.

Marjar's fighting story is well known to people on earth. So can you help to calculate the possibility of that Marjar generates no more than N organs in one day?

#### Input

The first line contains a single integer T (0 ≤ T ≤ 10000), indicating there are T cases in total. Then the following T lines each contains one integer N (1 ≤ N ≤ 100) and one float number λ (1 ≤ λ ≤ 100), which are described in problem statement.

#### Output

For each case, output the possibility described in problem statement, rounded to 3 decimal points.

#### Sample Input

3
5 8.000
8 5.000
2 4.910

#### Sample Output

0.191
0.932
0.132
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <map>
#include <queue>
#include <sstream>
#include <iostream>
using namespace std;
#define INF 0x3fffffff

int main()
{
//freopen("//home//chen//Desktop//ACM//in.text","r",stdin);
//freopen("//home//chen//Desktop//ACM//out.text","w",stdout);
int T;
scanf("%d",&T);
while(T--)
{
int n;
double l;
scanf("%d%lf",&n,&l);
double sum=0;
for(int k=0;k<=n;k++)
{
double tmp;
tmp=exp(-l);
for(int i=k;i>=1;i--)
{
tmp=tmp*l;
tmp=tmp/i;
}
sum+=tmp;
}
printf("%.3lf\n",sum);
}
return 0;
}