Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 260 Accepted Submission(s): 187
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=1e3+10;
ll gcd(ll a,ll b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
int main()
{
ll n,m;
while(~scanf("%lld%lld",&n,&m)&&(n||m))
{
ll fenzi=n*(m-1),fenmu=m*m;
while(1)
{
ll k=gcd(fenzi,fenmu);
if(k==1) break;
fenzi/=k;fenmu/=k;
}
printf("%lld/%lld\n",fenzi,fenmu);
}
return 0;
}
分析:比赛时就感觉是个什么分布,,但是学的很多又忘了,最后百度了一下,才发现可以二项分布做;
对于每个盒子,每个球落入其中的概率是p=1/m;
那么总共n个球p(x=k)=C(n,k)*p^k*(1-p)^(n-k),显然的二项分布;
二项分布数学期望E(x)=np(n是实验次数,p是每次试验球落入盒子的概率);
方差D(x)=np(1-p)
本题中D(x)=n/m*(1-1/m)=n*(m-1)/(m^2);
原文:http://www.cnblogs.com/smilesundream/p/5755971.html