HDU 5810 Balls and Boxes
Balls and Boxes
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 291 Accepted Submission(s): 213
Problem Description
Mr. Chopsticks is interested in random phenomena, and he conducts an experiment to study randomness. In the experiment, he throws n balls into m boxes in such a manner that each ball has equal probability of going to each boxes. After the experiment, he calculated the statistical variance V as
where Xi is the number of balls in the ith box, and X¯ is the average number of balls in a box.
Your task is to find out the expected value of V.
V=∑mi=1(Xi−X¯)2m
where Xi is the number of balls in the ith box, and X¯ is the average number of balls in a box.
Your task is to find out the expected value of V.
Input
The input contains multiple test cases. Each case contains two integers n and m (1 <= n, m <= 1000 000 000) in a line.
The input is terminated by n = m = 0.
The input is terminated by n = m = 0.
Output
For each case, output the result as A/B in a line, where A/B should be an irreducible fraction. Let B=1 if the result is an integer.
Sample Input
2
1
2
2
0
0
Sample Output
0/1
1/2
Hint
In the second sample, there are four possible outcomes, two outcomes with V = 0 and two outcomes with V = 1.
球落入每个盒子的概率是1/m
概率不变,n个球 可以当作n次实验 可用二项分布做
方差 D(x)=n*p*q
p=1/m q=1-1/m
带入可得 D(x)=n*(m-1)/(m*m)
方差的期望就是 方差的均值 而每个盒子的方差是相等的,所以期望的值就是方差的值了
/* *********************************************** Author :guanjun Created Time :2016/8/10 12:46:03 File Name :hdu5810.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 0x3f3f3f3f #define maxn 10010 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; priority_queue<int,vector<int>,greater<int> >pq; struct Node{ int x,y; }; struct cmp{ bool operator()(Node a,Node b){ if(a.x==b.x) return a.y> b.y; return a.x>b.x; } }; bool cmp(int a,int b){ return a>b; } int main() { #ifndef ONLINE_JUDGE //freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); ll n,m; while(scanf("%lld %lld",&n,&m)!=EOF){ if(n==0&&m==0)break; ll a=n*(m-1); ll b=m*m; ll k=__gcd(a,b); printf("%lld/%lld\n",a/k,b/k); } return 0; }
原文地址:http://www.cnblogs.com/pk28/
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