HDU-Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 285219 Accepted Submission(s): 67716

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 
     
  

本来想用常规方法做的，结果发现超时了。。。

    for(int i=1;i<=n;i++){
      int x;
      scanf("%d",&x);
      a[i]=a[i-1]+x;
      for(int j=0;j<i;j++){
        if(max1<a[i]-a[j]){
          max1=a[i]-a[j];
          ue=i;
          ub=j;
        }
    }
  }

因为数据范围是1e5，所以用二重循环会超时，然后在网上看到用dp写的

感觉很巧妙

  int k=1,s=0,e=0,summax=-1000,sum=0;  
      
        for(i=0;i<n;i++)  
        {  
            scanf("%d",&a[i]);  
        }  
        for(i=0;i<n;i++)  
        {  
            sum+=a[i];  
            if(sum>summax)  
            {  
                summax=sum;  
                s=k;  
                e=i+1;  
            }  
            if(sum<0)  
            {  
                sum=0;  
                k=i+2;  
            }  
        }

因为是连续的，所以用一个sum来记录连续的和

如果sum>0的话，与summax比较记录开始，结束位置

而如果sum<0的话，那么如果用这个sum与后面相加的话，肯定很亏，所以就吧sum赋值为0，把开始点变成下一位

嗯，很巧妙。。。。

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