hdu1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 209425    Accepted Submission(s): 49082


[align=left]Problem Description[/align]
Given a sequence a[1],a[2],a[3]......a
, 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.

 

[align=left]Input[/align]
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).

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

 

[align=left]Sample Output[/align]

Case 1:
14 1 4

Case 2:
7 1 6

代码：

#include <iostream>

#include<string.h>

using namespace std;

int a[100005];

int main()

{

    int t;

    cin>>t;

    for(int i=1;i<=t;i++)

    {

        int n;

        cin>>n;

        int s=1,ns=1,e=1,sum=-10002,max=-10002;    //s:start,e:end,ns:此时判断的起点 

       // memset(a,0,sizeof(a));

        for(int j=1;j<=n;j++)

        {

            cin>>a[j];

            if(sum+a[j]<a[j])

            {

                sum=a[j];

                ns=j;

            }

            else

                sum+=a[j];

        if(sum>max)

        {

            max=sum;

            s=ns;

            e=j;

        }

        }

        cout<<"Case "<<i<<":"<<endl;

        cout<<max<<" "<<s<<" "<<e<<endl;

        if(i!=t)

            cout<<endl;

    }

    return 0;
}

题意：求出和最大的序列

对于每个数有两种处理方式：

1.加到前面的序列中，变为前面最大和中的一个量

2.自己独立形成一个最大序列和

若前一个子段i-1的最大和<0，则最大子段的计数从i开始算，即选择第二种处理方式。