poj2479 Maximum sum

Maximum sum

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 37096 Accepted: 11570

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<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
/*代码一：TLE
int main()
{
int a[50001];
int left[50001];
int right[50001];
int T;
int n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
//left[i]表示i（包括i）左边最大字段和
left[0]=a[0];
for(int i=1;i<n;i++)
{
if(left[i-1]<0)
left[i]=a[i];
else
left[i]=left[i-1]+a[i];
}
//right[i]表示i（包括i）右边最大字段和
right[n-1]=a[n-1];
for(int i=n-2;i>=0;i--)
{
if(right[i+1]<0)
right[i]=a[i];
else
right[i]=right[i+1]+a[i];
}

int temp=-10000;
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
if(temp<(left[i]+right[j]))
temp=left[i]+right[j];
}
for(int i=0;i<n;i++)
if(temp<(left[i]+right[i+1]))
temp=left[i]+right[i+1];
printf("%d",temp);
}
return 0;
}
*/

//只接受O（N）的算法

int main()
{
int a[50001];
int left[50001];
int right[50001];
int T;
int n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
//left[i]表示i（包括i）左边最大字段和
left[0]=a[0];
for(int i=1;i<n;i++)
{
if(left[i-1]<0)
left[i]=a[i];
else
left[i]=left[i-1]+a[i];
}
//left[i]表示i左边的最大字段和
for(int i=1;i<n;i++)
{
left[i]=max(left[i],left[i-1]);
}
//right[i]表示i（包括i）左边最大字段和
right[n-1]=a[n-1];
for(int i=n-2;i>=0;i--)
{
if(right[i+1]<0)
right[i]=a[i];
else
right[i]=right[i+1]+a[i];
}
//right[i]表示i 右边最大字段和
for(int i=n-2;i>=0;i--)
right[i]=max(right[i],right[i+1]) ;
int temp=-10000;
for(int i=0;i<n-1;i++)
if(temp<(left[i]+right[i+1]))
temp=left[i]+right[i+1];
printf("%d\n",temp);
}
return 0;
}