NanoApe Loves Sequence-待解决

NanoApe Loves Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 440    Accepted Submission(s): 205


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.

1T10, 3n100000, 1Ai109
 

 

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
1 4 1 2 3 4
 

 

Sample Output
6
 

 

Source
 

 

Recommend
wange2014   |   We have carefully selected several similar problems for you:   5808  5807  5806  5805  5804 
 
#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
    int t;
    int n;
    int cha=0;
    int cha2=0;
    int a[100005]={0};
    int maxx1;
    int first;
    int maxx2;
    int second;
    int maxx3;
    int third;
    int sum=0;
    scanf("%d",&t);
    for(int z=0;z<t;z++){
        sum=0;
        maxx1=0;
        maxx2=0;
        maxx3=0;
        scanf("%d",&n);

        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            if(i!=0){
                cha=abs(a[i]-a[i-1]);
                if(maxx1<cha){
                    maxx1=cha;
                    first=i;
                }
            }
        }

        for(int i=1;i<n;i++){
            cha=abs(a[i]-a[i-1]);
            if(maxx2<cha){
                if(maxx2<=maxx1){
                    if(i==first){
                        continue;
                    }else{
                        maxx2=cha;
                        second=i;
                    }
                }
            }
        }

        for(int i=1;i<n;i++){
            cha=abs(a[i]-a[i-1]);
            if(maxx2<cha){
                if(maxx3<=maxx1&&maxx3<=maxx2){
                    if(i==first||i==second){
                        continue;
                    }else{
                        maxx3=cha;
                        third=i;
                    }
                }
            }
        }

        for(int i=1;i<n-1;i++){
            cha2=abs(a[i+1]-a[i-1]);
            if(maxx1<=cha2){
                sum+=cha2;
            }
            if(maxx1>cha2){
                if(i==first&&i+1==second||i==second&&i+1==first){
                    if(maxx3!=0){
                        if(maxx3>=cha2){
                            sum+=maxx3;
                        }else{
                            sum+=cha2;
                        }

                    }else{
                        sum+=cha2;
                    }

                }
                if(i==first||i==first-1){
                    if(cha2<=maxx2){
                        sum+=maxx2;
                    }else{
                        sum+=cha2;
                    }
                }else{
                    sum+=maxx1;
                }

            }
        }
        if(first==1){
            sum+=maxx2;
            sum+=maxx1;
        }
        if(first==n-1){
            sum+=maxx2;
            sum+=maxx1;
        }
        if(first!=1&&first!=n-1){
            sum+=(2*maxx1);
        }
        printf("%d\n",sum);
    }
    return 0;
}

 

 
 
全部评论

相关推荐

06-28 22:48
已编辑
广东金融学院 Java
小浪_Coding:学院本+这俩项目不是buff叠满了嘛
点赞 评论 收藏
分享
不要停下啊:大二打开牛客,你有机会开卷了,卷起来,去找课程学习,在牛客上看看大家面试笔试都需要会什么,岗位有什么需求就去学什么,努力的人就一定会有收获,这句话从来都经得起考验,像我现在大三了啥也不会,被迫强行考研,炼狱难度开局,啥也不会,找工作没希望了,考研有丝丝机会
点赞 评论 收藏
分享
评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客网在线编程
牛客网题解
牛客企业服务