POJ 2479 Maximum sum (同POJ 2593 Max Sequence)

Maximum sum

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 30031 Accepted: 9177

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

1101 -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.

Source

POJ Contest,Author:Mathematica@ZSU

 

代码：

#include<stdio.h>#include<string.h>#include<stdlib.h>#define MAX 50002#define NEG_INF -999999999int a[MAX],sum1[MAX],sum2[MAX];int main(){int t,n,i,ans;int s1,s2,max1,max2;while(scanf("%d",&t)!=EOF){while(t--){scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);sum1[i]=sum2[i]=NEG_INF;}ans=s1=s2=max1=max2=NEG_INF;for(i=0;i<n-1;i++){if(s1>0) s1+=a[i];else s1=a[i];if(max1<s1) max1=s1;sum1[i]=max1;if(s2>0) s2+=a[n-i-1];else s2=a[n-i-1];if(max2<s2) max2=s2;sum2[n-i-1]=max2;}for(i=0;i<n-1;i++)ans=ans>(sum1[i]+sum2[i+1]) ? ans : (sum1[i]+sum2[i+1]);printf("%d\n",ans);}}return 0;}

思路：

求最大字段和的变形，若分成两段一遍遍求会超时，注意优化方法