求解 -------------最大子段和问题

 1.通过分治法求解最大子段和问题

 递归来求解：

#include<stdio.h>
#define n0 10

int maxsum(int a[n0],int left,int right){
	int j,sum;
	int center,leftsum,rightsum;
	int s1,s2,lefts,rights;
	if(left==right) { sum=a[left];return sum;} //出口 
	    center=(left+right)/2;   
	    leftsum=maxsum(a,left,center);
		rightsum=maxsum(a,center+1,right);
		 s1=0; lefts=0;
		 for(j=center;j>=left;j--){
		 lefts=lefts+a[j];
		  if(lefts>s1)  s1=lefts;	
             }
    s2=0;rights=0;
    for(j=center+1;j<=right;j++){
    	rights=rights+a[j];
    	if(lefts>s2)  s2=rights;
	}
	sum=s1+s2;
	if(sum<leftsum)  sum=leftsum;
	if(sum<rightsum) sum=rightsum;
	return sum;
}

void main(){
	int a[n0]={0,5,5,5,5,5,0,-2,10,0};
	int left=0,right=n0-1,sum;
	int j;
    sum=maxsum(a,left,right);
    printf("序列（");
    for(j=0;j<n0;j++)
        printf("%5d",a[j]);
    printf("）\n");
	printf("的最大子段和为：%5d\n",sum); 
}

2.第二种就是直接枚举，暴利求解最大子问题，看着蛮简单的

#include<stdio.h>
#define n0 10
void main(){
	int a[n0]={5,5,5,5,5,0,0,0,-20,0};
	int sum,lsum;
	int i,j;
	sum=0;
	for(i=0;i<n0;i++)
	{
		lsum=0;
		for(j=i;j<n0;j++){
			lsum=lsum+a[j];
			if(lsum>sum) sum=lsum;
		}
	}
	printf("序列（");
	for(j=0;j<n0;j++)
	   printf("%5d",a[j]);
	printf(")\n");
	printf("的最大子段和为：%5d\n",sum);
}