第一次參加公司的招聘筆試，雖然只是抱著試試水的心態去參加的，可惜的是第一題就做錯了。。。。。 第一題，其實只是一個求最大子段和的變式題，不過筆試的時候也不知道怎麼了，就是不知道思路，最後還寫了一個錯的思路 題目大意：筆試題是求兩個不相鄰區間的最大子段和， OJ題是不相交區間的最大子段和
思路：很簡單，遍歷一遍，記錄從前往後每個元素以該元素結尾的最大子段和，從後往前每個元素以該元素結尾的最大欄位和， 然後一層遍歷列舉每一個斷點，求出該斷點前的最大子段和和斷點後的最大子段和最大值

Maximum sum

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 41980	Accepted: 13098

Description

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

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

Hint

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

Huge input,scanf is recommended. Maximum sum

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 41980	Accepted: 13098

Description

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

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

Hint

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

Huge input,scanf is recommended.

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<sstream>
#include<cctype>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI=acos(-1.0);
const double eps=1e-6;
const int INF=0x3f3f3f3f;
const int maxn=1234;

int T;
int a[50005];
int b1[50005],b2[50005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);

        int sum=0;
        int maxs=a[0];
        for(int i=0;i<n;i++)、、從前往後的最大子段和
        {
            sum+=a[i];
            if(sum>maxs) maxs=sum;
            if(sum<0) sum=0;

            b1[i]=maxs;
        }
        sum=0;
        maxs=a[n-1];
        for(int i=n-1;i>=0;i--){//從後往前的最大子段和
            sum+=a[i];
            if(sum>maxs) maxs=sum;
            if(sum<0) sum=0;
            b2[i]=maxs;
        }

       

        int ans=b1[0]+b2[1];
        for(int i=0;i<n-1;i++)
        {
            ans=max(ans,b1[i]+b2[i+1]);
        }

        printf("%d\n",ans);



    }

    return 0;
}