hud1003 max num

一道动态规划  F（x ） = max (F[x-1] + a[x] ,a[x])

题目：

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 123249    Accepted Submission(s): 28497

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

 

 

代码：

 

#include <iostream>
using namespace std;
const int maxn = 100000+10;
int arr[maxn];
int sum[maxn];
int main()
{
    int tst;
    cin>>tst;
    int cse= 0;
    while(tst--)
    {
        int n;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>arr[i];
        }

        int s=0,e=0,pos=0;
        int max = arr[0];
        int now  = arr[0];
        for(int i=1;i<n;i++)
        {
            if(now+arr[i]<arr[i])
            {
                pos = i;
                now = arr[i];
            }
            else
            {
                    now+=arr[i];

            }
            if(now > max)
            {
                s= pos;
                e = i;

                max = now;
            }
        }
        cout<<"Case "<<++cse<<":"<<endl;

        cout<<max<<" "<<s+1<<" "<<e+1<<endl;
        if(tst)
        cout<<endl;

    }


    return 0;
}