HDU-1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 231465 Accepted Submission(s): 54549

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

Author
Ignatius.L
解题思路：此题是一个动态规划的问题，关键的地方的判断前面数之和相加加上此数之和是否小于此数，如果小于，则需要改变起始位置，这也是找局部最大值，找到之后与此前保存的max进行比较，如果大于max则改变max,最终位置等于此数的位置。
解题代码

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
using namespace std;
int main()
{
    int T;
    scanf("%d",&T);
    int count1=1;
    while(T--)
    {

        int now,before,max1;
        int N;
        int s,e,x;
        scanf("%d",&N);
        for(int i=1;i<=N;i++)
        {
          scanf("%d",&now);
          if(i==1)
          {
              before=max1=now;
              x=s=e=1;
              continue;
          }
          if(now>now+before)
          {
              before=now;
              x=i;
          }
          else
          {
             before=before+now;
          }
          if(before>max1)
          {
              max1=before;
              s=x;
                e=i;
          }
        }
        printf("Case %d:\n",count1);
        cout<<max1<<" "<<s<<" "<<e<<endl;
        count1++;
        if(T==0)
            continue;
        printf("\n");
    }
    return 0;
}