HDU 5753 Permutation Bo

最新推荐文章于 2016-08-03 12:24:46 发布

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该博客主要探讨了如何解决HDU 5753题目，通过计算期望来解决问题。文章详细分析了两种情况——边缘和中间的概率，为读者提供了清晰的解题思路。

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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 function

f(h)=∑ni=1ci[hi>hi−1 and hi>hi+1]

Bo have gotten the value of

c1∼cn, 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(1≤n≤1000), second line contains

n non-negative integer

ci(0≤ci≤1000).

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

10−4, your solution will be accepted.

Sample Input

4
3 2 4 5
5
3 5 99 32 12

Sample Output

6.000000
52.833333

算期望，反正只有边缘和中间两种，计算一下概率即可。

#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int mod = 1e9 + 7;
const int N = 1e3 + 10;
const int INF = 0x7FFFFFFF;
int T, n, c[N];
int f[N];

int main()
{
    //scanf("%d", &T);
    while (scanf("%d", &n) != EOF)
    {
        rep(i, 1, n) scanf("%d", &c[i]);
        double ans = (c[1] + c[n]) / 2.0;
        double res = 0;
        rep(i, 3, n) res += (i - 1)*(i - 2);
        ((res /= n) /= n - 1) /= n - 2;
        rep(i, 2, n - 1) ans += res*c[i];
        printf("%.6lf\n", ans);
    }
    return 0;
}