Maximum sum

                  Maximum sum

 

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u  

 

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.

 

解析:

大意：求两个字段最大和

分析：O(n^3)定超时，需优化

# include<stdio.h>
# include<string.h>
int dp1[50002];
int dp2[50002];
int s[50002];
int sign[50002];
int max(int a,int b)
{
    return a>b?a:b;
} 
int main()
{
   int nCase,nNum,i,Max;
   scanf("%d",&nCase);
   while(nCase--)
   {
       Max=-99999;
       scanf("%d",&nNum);
       dp1[0]=0;
       for(i=1;i<=nNum;i++)
       {
           scanf("%d",&s[i]);
       dp1[i]=max(s[i],dp1[i-1]+s[i]);
       }
       dp2[nNum]=s[nNum];
         sign[nNum]=s[nNum];
       for(i=nNum-1;i>=1;i--)
       {
               dp2[i]=max(s[i],dp2[i+1]+s[i]);
               sign[i]=max(dp2[i],sign[i+1]);//优化标记，记录i后的最大字段和
       }
      
       for(i=1;i<nNum;i++)
       {
          /* int maxx=0;
           for(j=i+1;j<=nNum;j++)
           {
               if(maxx<dp2[j])
                   maxx=dp2[j];
           }*/
           if(Max<dp1[i]+sign[i+1])
           Max =dp1[i]+sign[i+1];
       }
       printf("%d\n",Max);
   }
   return 0;
}