POJ-2593 Max Sequence

Max Sequence

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16001		Accepted: 6715

Description

Give you N integers a1, A2 ... an (|ai| <=1000, 1 <= i <= N).

You should output S.

Input

The input would consist of several test cases. For each test case, one of the integer n (2 <= n <= 100000) is given on the first line. Second line contains N integers. The input is terminated by a, with N = 0.

Output

For each test of the input, print a line containing S.

Sample Input

5-5 9-5 11 200

Sample Output

40

Approximate test instructions: Give a series, find out the number of disjoint in the sequence of two sub-segments and, require sub-segment and maximum.

1#include <stdio.h>2 intMain ()3 {4     intleft[100005];5     intright[100005];6     inta[100005];7      while(1)8     {9         intn,i;Tenscanf"%d",&n); One         if(n==0) A              Break; -          for(i=0; i<n;i++) -scanf"%d",&a[i]); the         intSuml=0; -         intmax=-99999999; -          for(i=0; i<n;i++)//Maximum number of consecutive substrings from left to right to end of I//If this cycle does not understand, you can do the next hdu1003 -         { +suml=suml+A[i]; -             if(suml>max) +max=SumL; A             if(suml<0) atSuml=0; -left[i]=Max; -         } -max=-99999999; -         inttmp=-99999999; -         intSumr=0; in          for(i=n-1; i>=1; i--) -         { tosumr=sumr+A[i]; +             if(sumr>max) -max=Sumr; the             if(sumr<0) *Sumr=0; $             if(tmp<max+left[i-1])Panax Notoginsengtmp=max+left[i-1]; -          } theprintf"%d\n", TMP); +      } A      return 0; the}

POJ-2593 Max Sequence