3查找最长和的数字串

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.<br>

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).<br>

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.<br>

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Case 1:
14 1 4

Case 2:
7 1 6
思路：给出一串数字，求出他的子字串和最大是多少。
累加，一直到他之前的数小于零就算是断串，再从断串处开始搜索，继续以上步骤。
例如 5 6 -1 4 5 -7
      6  5 9 14 7  此时14就是最大值 结果就为14 1 4.
三个步骤：
1，初始化，将now before max赋值为第一个数，x动态的指针，最终付给start   end是结束地址
2，判断是否断串
    2.1   假如断串(判断是否断的条件是 now > now + before)，从断串处重新开始搜索此时before = now，x=i；
    2.2   不断串，累加，before = before + now;
3,最大值 如果before > max 则max = before,并且 start = x;end = i；
代码：
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int t,n,now,before,start,eend,flot = 1,x,mmax;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;++i)
        {
            scanf("%d",&now);
            if(i==1)//初始化
            {
                before = now;
                mmax = now;
                x = 1;//动态指针，用来预存初始位置，最后赋予start
                start = 1;//初始位置
                eend = 1;//结束位置
            }
            else//开始计算
            {
                if(now > now + before)//断串的情况
                {
                    before = now;//从此处开始搜索
                    x = i;//初始位置变化啦
                }
                else
                    before = before + now;//没有断的话，一直进行累加
                if(before > mmax)//比之前的before大了
                {
                    start = x;
                    eend = i;
                    mmax = before;
                }
            }


        }
        printf("Case %d:\n%d %d %d\n",flot++,mmax,start,eend);
        //printf("Case %d:\n%d %d %d\n",flot++,mmax,start,eend);
        if(t)
            printf("\n");


    }
    return 0;
}