HDU 5753 Permutation Bo (组合数学)
来源:互联网 发布:swagger codegen java 编辑:程序博客网 时间:2024/04/29 06:52
题目链接:HDU 5753
题面:
Permutation Bo
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 376 Accepted Submission(s): 236
Special Judge
Problem Description
There are two sequences h1∼hn and c1∼cn .h1∼hn is a permutation of 1∼n . particularly, h0=hn+1=0 .
We define the expression[condition] is 1 when condition is True,is 0 when condition is False.
Define the functionf(h)=∑ni=1ci[hi>hi−1 and hi>hi+1]
Bo have gotten the value ofc1∼cn , and he wants to know the expected value of f(h) .
We define the expression
Define the function
Bo have gotten the value of
Input
This problem has multi test cases(no more than 12 ).
For each test case, the first line contains a non-negative integern(1≤n≤1000) , second line contains n non-negative integer ci(0≤ci≤1000) .
For each test case, the first line contains a non-negative integer
Output
For each test cases print a decimal - the expectation off(h) .
If the absolute error between your answer and the standard answer is no more than10−4 , your solution will be accepted.
If the absolute error between your answer and the standard answer is no more than
Sample Input
43 2 4 553 5 99 32 12
Sample Output
6.00000052.833333
Source
2016 Multi-University Training Contest 3
题意:
给定一个序列C,其长度为N,随后随机生成一个1-N的全排列,序列两端各有一个0,若序列中某位置比两端的值都大,则该位置对应的C数组的值,可以取得,并加入总值,问最后总值的期望。
解题:
比赛的时候,是队友猜的,也算是找规律吧。看了题解,说是6种情况,不是很明白,对于一个位置来说,不就左大又大,左小右小,左大右小,左小右大四种情况吗?看了网上十多篇题解,全都说找规律......也是醉了,唯独一篇讲的比较好,对于不是边缘的点,应该考虑其左右两侧和其自身三个位置的关系,全排列为6!,而符合的情况有2种,即左大右大(1),左大右大(2),即大的两个数位置可以交换,这点是没有考虑到的,故而概率为1/3,而对于边缘点,因为一个是0,故而概率为1/(2!)=1/2。随后,各位的值乘以概率的累加即为所求期望。
代码:
#include <iostream>#include <cstdio>using namespace std;int a[1005];int main(){int n;double ans=0;while(~scanf("%d",&n)){ans=0.0;for(int i=0;i<n;i++)scanf("%d",&a[i]); ans=(a[0]+a[n-1])/2.0;for(int i=1;i<n-1;i++)ans+=a[i]/3.0;printf("%.6lf\n",ans);}return 0;}
0 0
- HDU 5753 Permutation Bo (组合数学)
- 【数学】HDU 5753 Permutation Bo
- hdu 5753 Permutation Bo
- HDU 5753 Permutation Bo
- HDU-5753-Permutation Bo
- HDU 5753 Permutation Bo
- 【HDU】5753 Permutation Bo
- hdu5753 Permutation Bo(数学)
- hdu 5753 Permutation Bo (水题)
- HDU 5753 Permutation Bo (找规律)
- HDU-5753-Permutation Bo(找规律)
- MUlti 2016 Permutation Bo(hdu 5753)
- HDU 5753 Permutation Bo(期望)
- HDU 5753 Permutation Bo(水~)
- HDU-5753 Permutation Bo(期望)(概率)
- hdu 5753 Permutation Bo(2016 Multi-University Training Contest 3——组合)
- hdu 5225 Tom and permutation(组合数学)
- [HDU 5753] Permutation Bo (期望的线性性质)
- CSU 1100: 一二三
- linux 获取某个日期对应的月末日期
- HTTP缓存控制小结
- Makefile 变量$@,$^,$<的区别
- 服务器TIME_WAIT和CLOSE_WAIT详解和解决办法
- HDU 5753 Permutation Bo (组合数学)
- nagios系列(六)之nagios实现对服务器cpu温度的监控
- mac下升级ruby环境版本
- php url重写
- Word文件写着写着就变成了只读的问题解释
- Python自定义包引入
- LeetCode---6.RestEasy
- HTML5 codecademy table表格
- 文件搜索命令