hdu5810Balls and Boxes+数学期望
2016-08-16 21:18
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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
V=∑mi=1(Xi−X¯)2/m
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.
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.
Author
SYSU
Source
2016 Multi-University Training Contest 7
就是把n个球扔到m个箱子里。扔到每个箱子的概率是相等的,求箱子里球的个数的方差的期望。
这是个随机概率。每个箱子的概率相等,那么方差的期望就是总体的方差。
s=n*p*(1-p);
球等概率的扔到箱子里,所以p=1/m;
所以s=n∗(m−1)/m2
还有一个详细的推导过程(然而我都不会)
然而某神打表,观察出了规律。秒过。。然而并没有要到打表代码。。。
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
V=∑mi=1(Xi−X¯)2/m
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.
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.
Author
SYSU
Source
2016 Multi-University Training Contest 7
就是把n个球扔到m个箱子里。扔到每个箱子的概率是相等的,求箱子里球的个数的方差的期望。
这是个随机概率。每个箱子的概率相等,那么方差的期望就是总体的方差。
s=n*p*(1-p);
球等概率的扔到箱子里,所以p=1/m;
所以s=n∗(m−1)/m2
#include<cstdio> #include<string> #include<cmath> #include<algorithm> #include<iostream> using namespace std; #define LL long long int main(){ LL n,m; while(scanf("%I64d %I64d",&n,&m)!=EOF){ if(n==0&&m==0) break; LL fz=n*(m-1); LL fm=m*m; LL gcd=__gcd(fz,fm); fz=fz/gcd; fm=fm/gcd; printf("%I64d/%I64d\n",fz,fm); } return 0; }
还有一个详细的推导过程(然而我都不会)
然而某神打表,观察出了规律。秒过。。然而并没有要到打表代码。。。
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