Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 206582    Accepted Submission(s):
48294

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

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题意：求连续子序列和的最大值。

#include <string.h>
#include <iostream>
using namespace std;
int main()
{
    int i,n,t,start,end,c=1,a[100005];
    cin>>t;
    while (t--)
    {
        cin>>n;
        start=0;end=0;
        int max=-9999,k=0,sum=0;
        for (i=0;i<n;i++)
        {
            cin>>a[i];
            sum+=a[i];
            if (sum>max)
            {
                max=sum;
                start=k+1;
                end=i+1;
            }
            //cout<<i+1<<"       "<<sum<<" "<<max<<endl;
            if (sum<0) //0的意义就是这段数做的是负功
            {
                k=i+1;
                sum=0;
            }
        }
        cout<<"Case "<<c++<<":"<<endl;
        cout<<max<<" "<<start<<" "<<end<<endl;
        if (t)
        cout<<endl;
    }
    return 0;
}