Max Sum（杭电OJ1003）解题报告

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

解题思路：刚开始用最笨的方法试了一下，暴力搜索，全部走一边，时间花太多

了，超时了，对此改进了搜索的方式，只要两个循环就搞定了，时间加快了，但是相

对于来说还不是最快的，还是会超时的，一个个向后面走，只要和大于Max的，就记

录的，代码如下：

Code:

1. #include<iostream>
2. usingnamespacestd;
3. intmain()
4. {
5. intn;
6. cin>>n;
7. intpoint1,point2;
8. intstep=1;
9. while(n)
10. {
11. intmax=-99999;
12. intm,data[100000];
13. cin>>m;
14. for(inti=1;i<=m;i++)
15. cin>>data[i];
16. for(i=1;i<=m;i++)
17. {
18. intsum=0;
19. for(intj=i;j<m;j++)
20. {
21. sum+=data[j];
22. if(sum>max)
23. {
24. max=sum;
25. point1=i;
26. point2=j;
27. }
28. }
29. }
30. if(step!=1)
31. cout<<endl;
32. cout<<"Case"<<step<<":"<<endl;
33. cout<<max<<""<<point1<<""<<point2<<endl;
34. step++;
35. n--;
36. }
37. return0;
38. }

交上去还是超时的，最近用到动态规划，对算法进行改进了，起始点还是从头开始的，

一直到后面搜索，一直和为小于零，起始点就从开始小于零的后一位开始并把sum改为

零，再搜索的过程中，一遇到大的数据就记录下来，把其计为起始点和终点的，这里

面主要考虑到，当你搜索到一个位置的，它的和不小于零的，那对于后面来说，加上

去还是会变大的，不会给变小的，所以要再搜索下去的，走一边就KO了。代码如下：

Code:

1. #include<iostream>
2. usingnamespacestd;
3. #defineMin-999999
4. intmain()
5. {
6. intdata[100000],start,end;
7. intm;
8. intstep=1;
9. cin>>m;
10. while(m--)
11. {
12. intn;
13. cin>>n;
14. for(inti=1;i<=n;i++)
15. cin>>data[i];
16. intmax=Min;
17. intk=1;
18. intsum=0;
19. for(i=1;i<=n;i++)
20. {
21. sum=sum+data[i];
22. if(sum>max)
23. {
24. max=sum;
25. start=k;
26. end=i;
27. }
28. if(sum<0)
29. {
30. sum=0;
31. k=i+1;
32. }
33. }
34. if(step!=1)
35. cout<<endl;
36. cout<<"Case"<<step<<":"<<endl;
37. cout<<max<<""<<start<<""<<end<<endl;
38. step++;
39. }
40. return0;
41. }