NanoApe Loves Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)

Problem Description
NanoApe, the Retired Dog, has returned back to prepare for the National Higher Education Entrance Examination!

In math class, NanoApe picked up sequences once again. He wrote down a sequence with n numbers on the paper and then randomly deleted a number in the sequence. After that, he calculated the maximum absolute value of the difference of each two adjacent remained numbers, denoted as F.

Now he wants to know the expected value of F, if he deleted each number with equal probability.

Input
The first line of the input contains an integer T, denoting the number of test cases.

In each test case, the first line of the input contains an integer n, denoting the length of the original sequence.

The second line of the input contains n integers A1,A2,…,An, denoting the elements of the sequence.

1≤T≤10, 3≤n≤100000, 1≤Ai≤109

Output
For each test case, print a line with one integer, denoting the answer.

In order to prevent using float number, you should print the answer multiplied by n.

Sample Input

Sample Output
6

比賽的時候想到的是記錄前三大的值，還要分別考慮這三個值和去掉的數原本的左右差值和去掉之後的差值的大小關係，實現起來容易有坑，WA的一聲就哭了……

f1正序記錄前i個數中差值的絕對值的最大值，f2倒序記錄i後面的數中差值的絕對值的最大值。最後再從f1[i - 1]、f[i + 1]和abs(arr[i - 1] - arr[i + 1])中取最大值。

最後要加上去掉arr[0]和arr[n - 1]的情況。

PS:沒有試驗是否會爆int，直接寫的long long.

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define MAX 100005
#define mod 1000000007
#define INF -0x3f3f3f3f

using namespace std;

long long int arr[MAX];
long long int f1[MAX], f2[MAX];

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)  
            scanf("%I64d", &arr[i]);

        long long int ans = 0;
        memset(f1, 0, sizeof(f1));
        memset(f2, 0, sizeof(f2));
        for(int i = 1; i < n; i++)  
            f1[i] = max(f1[i - 1], abs(arr[i] - arr[i - 1]));
        for(int i = n - 2; i >= 0; i--) 
            f2[i] = max(f2[i + 1], abs(arr[i] - arr[i + 1]));

        for(int i = 1; i < n - 1; i++)  
            ans += max(max(f1[i - 1], f2[i + 1]), abs(arr[i - 1] - arr[i + 1]));
        ans += f1[n - 2];
        ans += f2[1];
        printf("%I64d\n", ans);
    }
    return 0;
}

執行結果：