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  • Codeforces_789C_(dp)

    C. Functions again
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:

    In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.

    Input

    The first line contains single integer n (2 ≤ n ≤ 10^5) — the size of the array a.

    The second line contains n integers a1, a2, ..., an (-10^9 ≤ ai ≤ 10^9) — the array elements.

    Output

    Print the only integer — the maximum value of f.

    Examples
    input
    5
    1 4 2 3 1
    output
    3
    input
    4
    1 5 4 7
    output
    6

    题意:求公式的最大值。
    思路:
      b[i]:abs(num[i]-num[i+1]),c[N]:-abs(num[i]-num[i+1])  (i%2==1)
      b[i]:-abs(num[i]+num[i+1]),c[N]:abs(num[i]-num[i+1])  (i%2==0)
      然后分别求b,c的区间连续最大值,二者中较大着就是答案。
    #include<bits/stdc++.h>
    using namespace std;
    #define N 100005
    #define LL long long
    #define INF 10000000005
    int num[N],b[N],c[N];
    
    int main()
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&num[i]);
        for(int i=0;i<n-1;i++)
        {
            if(i%2)
            {
                b[i]=abs(num[i]-num[i+1]);
                c[i]=-abs(num[i]-num[i+1]);
            }
            else 
            {
                b[i]=-abs(num[i]-num[i+1]);
                c[i]=abs(num[i]-num[i+1]);
            }
        }
        LL res1=-INF,res2=-INF,tmp=0,l=0;
        for(int i=0;i<n-1;i++)
        {
            tmp+=b[i];
            if(tmp>res1)
                res1=tmp;
            if(tmp<0)
                tmp=0;
        }
        tmp=0;
        for(int i=0;i<n-1;i++)
        {
            tmp+=c[i];
            if(tmp>res2)
                res2=tmp;
            if(tmp<0)
                tmp=0;
        }
        printf("%I64d
    ",max(res1,res2));
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/jasonlixuetao/p/6652904.html
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