[ACM]Max Sum

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

代码：

#include <iostream>
using namespace std;
int a[100002];//用来存放数字
int main()
{
    int T,N,i,j;
    cin>>T;
    for(i=1;i<=T;i++)
    {
        cin>>N;
        for(j=0;j<N;j++)
            cin>>a[j];
        int start=0,end=0,sum=0,s=0,e,max;//start起始位置，end结尾位置,sum为连续数字的和，s为起始位置变量，和start有关，e为结尾位置变量,max为所求的最大值
        max=a[0];
        for(j=0;j<N;j++)
        {
            e=j;//记录运算到得位置，也是末位置
            sum=sum+a[j];
            if(sum>max)
            {
                max=sum;
                start=s;
                end=e;
            }
            if(sum<0)//这里写成sum+a[j]<a[j]比较好理解,sum为负数，舍去前边的，从下一位开始从头运算
            {
                s=j+1;//记录新运算的起始位置
                sum=0;
            }
        }
        cout<<"Case "<<i<<":"<<endl;
        cout<<max<<" "<<start+1<<" "<<end+1<<endl;
        if(i<=T-1)
            cout<<endl;//注意格式

    }
    return 0;
}

运行截图：