1481:Maximum sum

Total time limit: 1000ms memory limit: 65536kB 如果有想看中文版的同学，轻击中文版，如果有想看源代码的同学，轻击源代码。

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

1 10 1 -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; } }