【noiOJ】p1481

1481:Maximum sum

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总时间限制: 

1000ms

内存限制: 

65536kB

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.

来源

POJ Contest,Author:Mathematica@ZSU

source:最大双子序列和，枚举i，分别统计i左边最大子序列和右边最大子序列。（改变一下代码风格）

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <cstring>
 using namespace std;
 int n,datanum,maxn;
 int a[+],leftmax[+],rightmax[+];
 int main()
 {
     int w,i;
     scanf("%d",&datanum);
     for (w=;w<=datanum;w++)
     {
         memset(a,,sizeof());
         memset(leftmax,,sizeof(leftmax));
         memset(rightmax,,sizeof(rightmax));
         scanf("%d",&n);
         for (i=;i<=n;i++)
             scanf("%d",&a[i]);
         leftmax[]=a[];
         for (i=;i<=n;i++)
             if (leftmax[i-]>)        leftmax[i]=leftmax[i-]+a[i];
             else    leftmax[i]=a[i];
         for (i=;i<=n;i++)
             if (leftmax[i]<leftmax[i-])    leftmax[i]=leftmax[i-];
         rightmax[n]=a[n];
         for (i=n-;i>=;i--)
             if (rightmax[i+]>)    rightmax[i]=rightmax[i+]+a[i];
             else    rightmax[i]=a[i];
         for (i=n-;i>=;i--)
             if (rightmax[i]<rightmax[i+])
                 rightmax[i]=rightmax[i+];
         maxn=-;
         for (i=;i<n;i++)
             maxn=max(maxn,leftmax[i]+rightmax[i+]);
         printf("%d\n",maxn);
     }
     return ;
 }