Balls and Boxes(盒子与球) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 Mr. Chopsticks is interested in random phenomena, and he conducts an experiment to study randomness. In the experiment, he thr…

Balls and Boxes 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5810 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 t…

Balls and Boxes 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5810 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 t…

Balls and Boxes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 260    Accepted Submission(s): 187 Problem Description Mr. Chopsticks is interested in random phenomena, and he conducts an experi…

n*(m-1)/(m*m) #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #include<set> #include<queue…

官方题解看不太懂,参考了一些人的博客以后自己证明如下: 其中D(X)和E(X)的公式如下(参考自百度百科): 其中 p = 1 / m .(这是每一个单独事件发生的概率期望,在这里单独事件指的是一个球放到m个盒子里面,放到每个盒子里面的概率都是一样的,所以期望都是1 / m了) 推出了公式以后,代码就很好写了.代码如下: #include <stdio.h> #include <algorithm> #include <string.h> using namespace…

题目链接 这题似乎就是纯概率论.. E(V)=D(X_i)=npq (p=1/m,p+q=1) #include<bits/stdc++.h> using namespace std; typedef long long LL; LL n,m; LL gcd(LL a,LL b) { return b?gcd(b,a%b):a; } int main() { while(cin>>n>>m && n+m) { LL a=n*m-n,b=m*m; LL g…

C. Balls and Boxes time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Vasya had n boxes with balls in the room. The boxes stood in a row and were numbered with numbers from 1 to n from…

Balls and Boxes                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)                                                                            …

260C - Balls and Boxes 思路:模拟.在x前面找到最小值,如果没有,从0跳到n,继续找到最小值,边找最小值路过的点边减1.然后所有值都减去最小值,最小值那个点加上减去的值. 找到x前面离x最近的最小值的原因:如果如果在x到最小值之间还有最小值,那么这个最小值最后会变成-1. 简单代码: #include<bits/stdc++.h> using namespace std; #define ll long long ; const ll INF=0x7f7f7f7f; ll…