【openjudge】Maximum sum

1481:Maximum sum

Total time limit: 1000ms
memory limit: 65536kB
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describe

Given a set of n integers: A={a1, a2,…, an}, we define a function d(A) as below:

                     t1     t2          d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n }                    i=s1   j=s2

Your task is to calculate d(A).

input

The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, …, an. (|ai| <= 10000).There is an empty line after each case.

output

Print exactly one line for each test case. The line should contain the integer d(A).

sample input

1101 -1 2 2 3 -3 4 -4 5 -5

sample output

13

Prompt

In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.

source

POJ Contest,Author:Mathematica@ZSU

【analysis】+【Code】

状态: Accepted

#include<cstring>#include<iostream>#include<algorithm>using namespace std;int n,nmax,a[50001],f[50001],g[50001],F[50001],G[50001];int main(){    int t;    cin>>t;    while(t--)    {        nmax=-10001;        cin>>n>>a[1];        int mxf=F[1]=f[1]=a[1];//统一开头        for(int i=2;i<=n;i++)//两段for，分别从前后检查        {            cin>>a[i];            f[i]=a[i]+max(0,f[i-1]);//加上前面一个的数，如果是负数，就不加            mxf=F[i]=max(mxf,f[i]);//储存最大f[]        }        int mxg=G[n]=g[n]=a[n];//统一g[]结尾           for(int i=n-1;i>0;i--)//不能包含n        {            g[i]=a[i]+max(0,g[i+1]);//储存最大（加上后一个数（如果不为负））            G[i]=max(mxg,g[i]);            mxg=G[i];        }        for(int i=1;i<n;i++)            if(F[i]+G[i+1]>nmax)                nmax=F[i]+G[i+1];        cout<<nmax<<endl;    }}