SGU 495 Kids and Prizes [期望]

Description

ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the company organized a contest for kids for the best design of a cardboard box and selected M winners. There are N prizes for the winners, each one carefully packed in a cardboard box (made by the ICPC, of course). The awarding process will be as follows:

• All the boxes with prizes will be stored in a separate room.
• The winners will enter the room, one at a time.
• Each winner selects one of the boxes.
• The selected box is opened by a representative of the organizing committee.
• If the box contains a prize, the winner takes it.
• If the box is empty (because the same box has already been selected by one or more previous winners), the winner will instead get a certificate printed on a sheet of excellent cardboard (made by ICPC, of course).
• Whether there is a prize or not, the box is re-sealed and returned to the room.

The management of the company would like to know how many prizes will be given by the above process. It is assumed that each winner picks a box at random and that all boxes are equally likely to be picked. Compute the mathematical expectation of the number of prizes given (the certificates are not counted as prizes, of course).

題意:

有N個箱子裏裝着禮物，M個人過來挑箱子，每個人是獨立的，當一個人選了一個有禮物的箱子，他會拿走禮物，並把空箱子放回去，也就是說下一個人選這個箱子的話會得不到禮物，問最終M個人得到禮物個數的期望。

範圍：

N<=10W

解法：

如果從最後會有幾個箱子被打開過這一角度來思考，這一題非常難做，至少是N^2的複雜度，所以需要轉換思想，計算每個禮物對最後答案的貢獻，可知每個人相互獨立，那麼可以計算得到每個箱子被M個人打開的概率P，然後n*p便是答案（二項分佈的期望公式）。

代碼：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;



int main(){
    int n,m;
    scanff(n);
    scanff(m);
    double p=double(n-1)/double(n);
    double t=double(n);
    rep(i,1,m){
        t=t*p;
    }
    double ans=double(n)-t;
    printf("%.10f\n",ans);

    return 0;
}