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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 164592 Accepted Submission(s): 38540

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

#include<iostream>
using namespace std;
int main()
{
    int T;//T為輸入例子的個數 
    cin>>T;
    for(int i=1;i<=T;i++)
    {
        int N;//N為每個例子中數字的個數 
        cin>>N;
        int s=0,start1=1
 
,start2,end,maxsum=INT_MIN;
        //s為以目前為止的點作為結束的一類中，最大的和 ；
        //start1為以目前為止的點作為結束的一類中最大和序列的起點坐標
        //start2為最終確定的最大和序列中的起點坐標
        //end為最終確定的最大和序列結束點的坐標
        //maxsum為最終確定的最大和
        for(int j=1;j<=N;j++)
        { 
           int a;
           cin>>a;
           if(s<0)
           {
                s=a;
                start1=j;
           }
           else
           {
                s+=a;
           }
           if(s>maxsum)
           {
                maxsum=s;
                start2=start1;
                end=j;
           }
        }
        cout<<"Case "<<i<<":"<<endl;
        cout<<maxsum<<" "<<start2<<" "<<end<<endl;
        if(i<T)
          cout<<endl;
    }
}

具體原理可參考http://alorry.blog.163.com/blog/static/6472570820123801223397/

HDU acm 1003 Max Sum || 動態規劃求最大子序列和詳解