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  • HDU 5753 Permutation Bo

    Permutation Bo

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 485    Accepted Submission(s): 286
    Special Judge


    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=ci[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
     
    Source
     
     
     
    解析:根据题意,可以考虑每个位置对期望的贡献。当i不在排列两端的时候,有3! = 6种大小关系,其中有2种对期望有贡献,因此贡献为ci/3;当i在排列两端时,有2! = 2种大小关系,其中有1种对期望有贡献,因此贡献为ci/2。据此即可解答本题(注意特判n == 1的情况)
     
     
     
    #include <bits/stdc++.h>
    
    int n;
    double c[1005];
    
    int main()
    {
        while(~scanf("%d", &n)){
            for(int i = 1; i <= n; ++i)
                scanf("%lf", &c[i]);
            if(n == 1){
                printf("%f
    ", c[1]);
                continue;
            }
            double ans = (c[1]+c[n])/2;
            for(int i = 2; i < n; ++i)
                ans += c[i]/3;
            printf("%f
    ", ans);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/inmoonlight/p/5710764.html
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