Find the largest sub-segment sum (divide and conquer)

Algorithm idea:

Find the largest sub-segment sum through the idea of ​​divide and conquer, and divide the number component into two parts equally, then the largest sub-segment sum will exist in three situations:
1. The largest sub-segment sum appears at the left end
2. The largest sub-segment sum appears At the right end
3. The largest sub-segment and straddling the left and right segments, the largest sub-segment sum is obtained by comparing the sizes

 

//最大子段和问题 p64 

#include<stdio.h>
#include<iostream>
using namespace std;
int MaxSum(int a[],int left,int right);
int a[100];
int main(void)
{
	int n,j;
	scanf("%d",&n);  
	for(j=0;j<n;j++)
	scanf("%d",&a[j]);	
	printf("%d",MaxSum(a,0,n-1));
	return 0;	
}

int MaxSum(int a[],int left,int right)
{
	int sum=0,midSum=0,leftSum=0,rightSum=0;
	int center,s1,s2,lefts,rights;
	if(left==right)
		sum=a[left];
	else{
		center=(left+right)/2;
		leftSum=MaxSum(a,left,center);   //情况1,最大字段和全部取左边元素 
		rightSum=MaxSum(a,center+1,right);//情况2,最大字段和全部取右边元素 
		s1=0;lefts=0;                      //情况3  最大子段和横跨中间 
		for(int i=center;i>=left;i--)      //求出s1:从中间到左边的最大和 
		{
			lefts+=a[i];
			if(lefts>s1) s1=lefts;
		}
		s2=0;rights=0;
		for(int j=center+1;j<=right;j++)    //求出s2 从中间到右边的最大和 
		{
			rights+=a[j];
			if(rights>s2) s2=rights;
		}
		midSum=s1+s2;              //横跨中间的最大字段和为s1+s2 
		if(midSum<leftSum) sum=leftSum;
		else sum=midSum;
		if(sum<rightSum) sum=rightSum;   //取三者的较大者		 
	}
	
	return sum;
	
	
}