題目鏈接：http://codeforces.com/problemset/problem/248/E
E. Piglet's Birthday
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Piglet has got a birthday today. His friend Winnie the Pooh wants to make the best present for him — a honey pot. Of course Winnie realizes that he won't manage to get the full pot to Piglet. In fact, he is likely to eat all the honey from the pot. And as soon as Winnie planned a snack on is way, the pot should initially have as much honey as possible.
The day before Winnie the Pooh replenished his honey stocks. Winnie-the-Pooh has n shelves at home, each shelf contains some, perhaps zero number of honey pots. During the day Winnie came to the honey shelves q times; on the i-th time he came to some shelf ui, took from it some pots ki, tasted the honey from each pot and put all those pots on some shelf vi. As Winnie chose the pots, he followed his intuition. And that means that among all sets of ki pots on shelf ui, he equiprobably chooses one.
Now Winnie remembers all actions he performed with the honey pots. He wants to take to the party the pot he didn't try the day before. For that he must know the mathematical expectation of the number m of shelves that don't have a single untasted pot. To evaluate his chances better, Winnie-the-Pooh wants to know the value m after each action he performs.
Your task is to write a program that will find those values for him.
Input
The first line of the input contains a single number n (1 ≤ n ≤ 105) — the number of shelves at Winnie's place. The second line contains nintegers ai (1 ≤ i ≤ n, 0 ≤ ai ≤ 100) — the number of honey pots on a shelf number i.
The next line contains integer q (1 ≤ q ≤ 105) — the number of actions Winnie did the day before. Then follow q lines, the i-th of them describes an event that follows chronologically; the line contains three integers ui, vi and ki (1 ≤ ui, vi ≤ n, 1 ≤ ki ≤ 5) — the number of the shelf from which Winnie took pots, the number of the shelf on which Winnie put the pots after he tasted each of them, and the number of the pots Winnie tasted, correspondingly.
Consider the shelves with pots numbered with integers from 1 to n. It is guaranteed that Winnie-the-Pooh Never tried taking more pots from the shelf than it has.
Output
For each Winnie's action print the value of the mathematical expectation m by the moment when this action is performed. The relative or absolute error of each value mustn't exceed 10 - 9.
Examples
input
Copy
3 2 2 3 5 1 2 1 2 1 2 1 2 2 3 1 1 3 2 2
output
Copy
0.000000000000 0.333333333333 1.000000000000 1.000000000000 2.000000000000
題意：
你有n(n<=1e5)個貨架，每個貨架上有a[i](a[i]<=100)罐蜂蜜，接下來要進行m(m<=1e5)次操作，每次操作格式爲
u v x (x<=5)
表示從第u個貨架隨機取出x罐蜂蜜進行品嚐，品嚐完後把這x罐蜂蜜全都放到第v個貨架上。（有這麼多蜂蜜就是任性）
現在問你，每次操作完後，期望的貨架個數，這些貨架上所有蜂蜜都被品嚐過。
思路：
顯然對於每個貨架，未被嘗過的蜂蜜罐數是單調不增的。
我們考慮用dp[i][j]表示第i個貨架上有j罐蜂蜜未品嚐過的概率。
對於輸入的a[i]顯然dp[i][a[i]]=1。初始答案ans爲所有dp[i][0]的和(1<=i<=n)
對於每次品嚐，我們枚舉品嚐完後該貨架上剩下的未被品嚐的蜂蜜罐數，再枚舉此次選的x罐中從未被品嚐過的罐數，用組合數計算一下概率，更新dp[i][j]的值。(1<=i<=n,0<=j<=a[i])
代碼：
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(register int i=(a);i<=(b);i++)
#define dep(i,a,b) for(register int i=(a);i>=(b);i--)
using namespace std;
const int maxn=1e5+5;
int n,m,k;
ll C[maxn][107];
ll a[maxn],b[maxn];
double dp[maxn][107],ans,tmp;
int main()
{
rep(i,0,100000)
{
C[i][0]=1;
rep(j,1,min(105,i))
C[i][j]=C[i-1][j]+C[i-1][j-1];
}
while(scanf("%d",&n)!=EOF)
{
ans=0;
rep(i,1,n) {scanf("%lld",&a[i]);b[i]=a[i];dp[i][a[i]]=1;}
rep(i,1,n) ans+=dp[i][0];
scanf("%d",&m);
while(m--)
{
int u,v,x;
scanf("%d%d%d",&u,&v,&x);
ans-=dp[u][0];
ll s=C[b[u]][x];
rep(i,0,a[u])
{
tmp=0;
rep(k,0,x)
if(b[u]>=i+k) tmp+=dp[u][i+k]*C[i+k][k]*C[b[u]-i-k][x-k]*1.0/(double)s;
dp[u][i]=tmp;
}
b[u]-=x; b[v]+=x;
ans+=dp[u][0];
cout<<fixed<<setprecision(12)<<ans<<endl;
}
}
return 0;
}