(HDU 5810)2016 Multi-University Training Contest 7 Elegant Construction (期望、方差、二项分布)
2016-08-09 22:57
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思路
咦,这不是方差么,要求样本方差的期望。。那不就是总体方差吗
可以发现把n个球放入m个里面,对于每一个球都是符合二项分布的(这里放或者不放)
期望定义:D=∑mi=1(Xi−X¯)2m
期望与方差的关系:D=E(X2)−E(X)2
二项分布的期望为:E=n×p,在题目中就是n÷m
二项分布的方差为:D=np(1−p)
其他分布的期望和方差:
代码
#include <bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define rep(i,a,b) for(int i=a;i<b;i++) #define debug(a) printf("a =: %d\n",a); const int INF=0x3f3f3f3f; const int maxn=1e5+50; const int Mod=1000000007; const double PI=acos(-1); typedef long long ll; using namespace std; int n,m; int main() { #ifndef ONLINE_JUDGE // freopen("in.txt","r",stdin); #endif //E :n/m //D=E(x^2)-E(x)^2 //E: np //D:np(1-p) while(scanf("%d %d",&n,&m)!=EOF && (n||m)){ ll fz=(ll)n*(m-1); ll fm=(ll)m*m; if (fz==0){ puts("0/1"); continue; } ll gc=__gcd(fz,fm); printf("%lld/%lld\n",fz/gc,fm/gc); } return 0; }
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