Solving the maximal sub-problems of sequences by divide-and-conquer algorithm

In particular, when all integers in a sequence are negative integers, their maximum is 0.

1#include <stdio.h>2 3 intCalumaxsubsum (int*array,intLeftintRight );4 5 intMain ()6 {   7     intarray[6] = {2, -2,3,1, -4,2};8     intLen_array =sizeof(array)/sizeof(array[0]);9     inti =0;Ten     intSubsum = calumaxsubsum (Array,0, Len_array); One      for(i =0; i < Len_array; i++) A     { -printf"%d", Array[i]); -     } the      -printf"\ n"); -     //printf ("Max Son and yes:"); -printf"subsum =%d\n", subsum);  +     return 0; - } +  A intCalumaxsubsum (int*array,intLeftintRight ) at {    -    intsum =0; -    if(Left >=Right ) -    { -       if(Array[left] >0) -       { insum =Array[left]; -       } to       Else +       {    -sum =0;  the       } *    } $    ElsePanax Notoginseng    { -       intCenter = (left + right)/2; the       intS1 =0, S2 =0; +       intLefts =0, rights =0; A       inti =0; the       intLeftsum =calumaxsubsum (array, left, center); +       intRightsum = calumaxsubsum (array, center +1, right); -        for(i = center; I >= left; i--) $       { $Lefts + =Array[i]; -          if(Lefts >S1) -          { theS1 =lefts; -          }Wuyi       } the        -        for(i = center +1; I < right; i++) Wu       { -Rights + =Array[i]; About           if(Rights >S2) $           { -S2 =Rights; -           }           -       } A        +sum = S1 +S2; the       if(Sum <leftsum) -       { $sum =leftsum; the       } the       if(Sum <rightsum) the       { thesum =rightsum; -       } in    } the     the    returnsum; About}

Where is the key idea?

Solving the maximal sub-problems of sequences by divide-and-conquer algorithm