杭电ACM1003C做法

Max Sum

Problem Description

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.

Input

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).

Output

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.

Sample Input

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

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

之前用java做这道题死活不对，最近又拿起了C，决定拿C做。自己的做法比较混乱，就贴上有注释的代码来吧！

#include

int main()
{
    int i,ca=1,t,s,e,n,x,now,before,max;
    scanf("%d",&t);
    while(t--)
    {
       scanf("%d",&n);
       for(i=1;i<=n;i++)
       {
         scanf("%d",&now);
         if(i==1)//初始化 
         {
            max=before=now;//max保留之前算出来的最大和，before存储目前在读入数据前保留的和，now保留读入数据  
            x=s=e=1;//x用来暂时存储before保留的和的起始位置，当before>max时将赋在s位置，s，e保留最大和的start和end位置 
         }
         else {
             if(now>now+before)//如果之前存储的和加上现在的数据比现在的数据小，就把存储的和换成现在的数据，反之就说明数据在递增，可以直接加上 
             {
                before=now;
                x=i;//预存的位置要重置 
             }       
             else before+=now;
              }
         if(before>max)//跟之前算出来的最大和进行比较，如果大于，位置和数据就要重置 
           max=before,s=x,e=i;
       }
       printf("Case %d:\n%d %d %d\n",ca++,max,s,e);
       if(t)printf("\n"); 
    }
    return 0;
}