A Maximum sum&&北邮月赛题-优快云博客

本文介绍了一种计算特定函数d(A)的方法，该函数应用于一组整数A中，任务是找到两组连续子序列，使得这两组子序列的元素之和最大。文章提供了一个高效的算法实现，并附带了示例输入输出，适用于算法竞赛或数据处理任务。

Time Limit:1000MS

Memory Limit:65536KB

Description

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

d(A)=sum{a[s1]~a[t1]}+sum{a[s2]~a[t2]}

The rule is 1<=s1<=t1<s2<=t2<=n.

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

Sample Output

Hint

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

Huge input,scanf is recommended

AC代码：

#include<iostream> #include<cstdio> #include<string.h> #define N 50001 using namespace std; int ss[N]; int main() { int T; scanf("%d",&T); while(T--) { int n,i; scanf("%d",&n); for(i=0;i<n;++i) scanf("%d",&ss[i]); int p=0; int s=0,e=0,t=0,max=0,sum=0; for(i=0;i<n;++i) { sum+=ss[i]; if(sum>max) { max=sum; s=t; e=i; } if(sum<0) { sum=0; t=i+1; } } p+=max; max=0; for(int i=s;i<=e;++i) ss[i]=0; for(int i=0;i<n;++i) { sum+=ss[i]; if(sum>max) max=sum; if(sum<0) sum=0; } printf("%d\n",p+max); }return 0; }