Maximum sum

描述

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).

输入

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.

输出

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

样例输入

1

10

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

样例输出

13

提示

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

Huge input,scanf is recommended.

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大意是说在给定的数列中，选取两段不重合的子串，使得其和最大。

写在前面：一个要注特别意的问题，WA的可能就是忽略了这个问题。

|ai| <= 10000，即可能有负数，那么初始化为0就不行了，0比负数大呀！怎么办呢，要么赋一个极小值，要么初始化为数列中的开始元。

还有一个小问题，呃，也不是什么问题，就是要注意一下数组的值的清空，因为即使定义在循环内部，他的生命周期结束了，可有可能下次分配到的地址还是在那里，里面的值并没有改变。

另外，对于多个字节的类型，memset只能初始化为 0 ，-1。不信的可以试试，一个整型四个字节，他会把每个字节都的二进制的每一位都赋值为目标数字……好像是这样，应该不对，反正不能赋值为其他值。

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#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int data[50005],l[50005],r[50005],mid[50005];
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        int n;
        scanf("%d",&n);
        for(int i = 1; i<=n; i++) {
            scanf("%d",&data[i]);
        }
        l[1] = data[1];
        for(int i = 2; i<=n; i++) {
            l[i] = 0;
            l[i] = max(data[i],l[i-1] + data[i]);
        }
        r[n] = data[n];
        for(int i = n-1; i>=1; i--) {
            r[i] = 0;
            r[i] = max(r[i+1] + data[i],data[i]);
        }
        mid[n] = r[n];
        for(int i = n - 1; i>=1; i--) {
            mid[i] = 0;
            mid[i] = max(r[i],mid[i+1]);
        }
        int ans = -1e9;
        for(int i = 2;i<=n;i++){
            ans = max(ans,l[i-1] + mid[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}