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

Input

The first line of the input contains an integer

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

Sample Input

1 4 1 2 3 4

Sample Output

Source

BestCoder Round #86

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#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;
}