Divide-and-conquer method (for maximal subsequence and)

1 //Find maximum sub-sequence 4, -3,5,-2,-1,2,6,-22#include <stdio.h>3 intMax (intAintBintc)4 {5     intret;6     if(A >b)7     {8RET =A;9}ElseTen     if(A <=b) One     { ARET =b; -     } -     if(Ret >=c) the     returnret; -     Else -     returnC; - } +  intFindmaxsum (intBox[],intSizeintLeftintright)//parameter (array name, array size, left border - { +     intMid = (right + left)/2; A     if(left = =Right )//Divide the recursion to pay attention to export conditions at     { -         returnBox[left]; -     } -     intLeftsum =findmaxsum (Box,size,left,mid); To find out the maximal sub-sequence of the left half and to have recursive trust, do not tangle layers in depth, assuming that the function is correct.  -     intRightsum = findmaxsum (Box,size,mid +1, right); The maximal sub-sequence of the right half region and -     intLeftbordersum =0; in     intRightbordersum =0; -     inti; to     intThissum =0; +      for(i = mid +1; I <= right;i++)//Find the maximal sub-sequence of the right half of the middle dividing point and (if the maximal sub-sequence crosses the intermediate dividing point, then it must contain the intermediate dividing point) -     { theThissum + =Box[i]; *         if(Rightbordersum <thissum) $         {Panax NotoginsengRightbordersum =thissum; -         } the      }  +Thissum =0; A       for(i = mid; I >= left;i--)//To find the maximal sub-sequence of the left half of the middle dividing point and the      { +Thissum + =Box[i]; -          if(Leftbordersum <thissum) $          { $Leftbordersum =thissum; -          } -      } the      intMidsum = Leftbordersum +rightbordersum; The maximal sub-sequence across the left and right half regions and -      returnMax (midsum,leftsum,rightsum); the largest sub-sequence of the left half and the largest sub-sequence of the right half, and the largest sub-sequence of the cross-section, and the largest of the three are the Seekers.Wuyi the      - } Wu intMain () - { About     intbox[8] = {4,-3,5,-2,-1,2,5,-2}; $     intRET =0; -ret = findmaxsum (box,8,0,7); -printf"%d", ret); -     return 0 ; A}

Legacy problem, this algorithm has a time complexity of O (NLOGN). think about how to find out.

Divide-and-conquer method (for maximal subsequence and)