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  • hdu-5753 Permutation Bo(概率期望)

    题目链接:

    Permutation Bo

    Time Limit: 2000/1000 MS (Java/Others)   

     Memory Limit: 131072/131072 K (Java/Others)

    Problem Description
    There are two sequences h1hn and c1cnh1hn is a permutation of 1n. particularly, h0=hn+1=0.

    We define the expression [condition] is 1 when condition is True,is 0 when condition is False.

    Define the function f(h)=ni=1ci[hi>hi1  and  hi>hi+1]

    Bo have gotten the value of c1cn, and he wants to know the expected value of f(h).
     
    Input
    This problem has multi test cases(no more than 12).

    For each test case, the first line contains a non-negative integer n(1n1000), second line contains n non-negative integer ci(0ci1000).
     
    Output
    For each test cases print a decimal - the expectation of f(h).

    If the absolute error between your answer and the standard answer is no more than 104, your solution will be accepted.
     
    Sample Input
    4
    3 2 4 5
    5
    3 5 99 32 12
     
    Sample Output
    6.000000
    52.833333
     
    题意:
     
    求这个函数值得期望;
     
    思路:
     
    发现首尾的概率为1/2,中间的为1/3;求一下就好;
     
    AC代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <bits/stdc++.h>
    #include <stack>
    
    using namespace std;
    
    #define For(i,j,n) for(int i=j;i<=n;i++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
    
    typedef  unsigned long long LL;
    
    template<class T> void read(T&num) {
        char CH; bool F=false;
        for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
        for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
        F && (num=-num);
    }
    int stk[70], tp;
    template<class T> inline void print(T p) {
        if(!p) { puts("0"); return; }
        while(p) stk[++ tp] = p%10, p/=10;
        while(tp) putchar(stk[tp--] + '0');
        putchar('
    ');
    }
    
    const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const int inf=1e9;
    const int N=1e7+10;
    const int maxn=1000+10;
    const double eps=1e-8;
    
    int c[1100];
    
    int main()
    {
            int n;
            while(cin>>n)
            {
                For(i,1,n)
                {
                    read(c[i]);
                }
                double ans=0;
                for(int i=2;i<n;i++)
                {
                    ans=ans+c[i]*1.0/3;
                }
                ans=ans+(c[1]+c[n])*1.0/2;
                printf("%.6lf
    ",ans);
            }
    
            return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5709008.html
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