连续子数组最大和

http://www.ahathinking.com/archives/tag/%E6%9C%80%E9%95%BF%E9%80%92%E5%A2%9E%E5%AD%90%E5%BA%8F%E5%88%97
求一个数组的相加和最大的连续子数组

思路：
一直累加，只要大于0，就说明当前的“和”可以继续增大，
如果小于0了，说明“之前的最大和”已经不可能继续增大了，就从新开始，
result=max{result+arr[i],arr[i]};//显然，若result>0，则可以继续相加，否则，就重新开始。

#include<stdio.h>
#define INF 65535

int Maxsum(int * arr, int size)
{
    int maxSum = -INF;
    int sum = 0;
    for(int i = 0; i < size; ++i)
    {
        if(sum < 0)
        {
            sum = arr[i];
        }else
        {
            sum += arr[i];
        }
        if(sum > maxSum)
        {
            maxSum = sum;
        }
    }
    return maxSum;
}


void Maxsum_location(int * arr, int size, int & start, int & end)
{
    int maxSum = -INF;
    int sum = 0;
    int curstart = start = 0;  /* curstart记录每次当前起始位置 */
    for(int i = 0; i < size; ++i)
    {
        if(sum < 0)
        {
            sum = arr[i];
            curstart = i;     /* 记录当前的起始位置 */
        }else
        {
            sum += arr[i];
        }
        if(sum > maxSum)
        {
            maxSum = sum;
            start = curstart; /* 记录并更新最大子数组起始位置 */
            end = i;
        }
    }
}


void main()
{
    /* 测试用例 */
    //int arr[] = {8,-10,3,60,-1,-6};
    int arr[] = {1,-2,3,5,-1,2};
    int arr2[] = {-9,-2,-3,-5,-4,-6};
    int len = sizeof arr / sizeof(int);
    int len2 = sizeof arr2 / sizeof(int);
 
    /* 测试实现 */
    printf("%d %d
",Maxsum(arr,len),Maxsum(arr2,len2));
 
    /* 返回起始位置 */
    int start = -1;
    int end = -1;
    Maxsum_location(arr,len,start,end);
    printf("start：%d end：%d
", start, end);

    Maxsum_location(arr2,len2,start,end);
    printf("start：%d end：%d
", start, end);
 
}