2015 HUAS Summer Trainning #5 A

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

 

题目大意：给你N个数，要你求最大的子段和（连续的）。

 

解题思路：

for(i=1;i<=n;i++)
        {
            cin>>d;
            t+=d;
            if(t>max)
            {
                start=k;
                end=i;
                max=t;
            }
            if(t<0)
            {
                t=0;
                k=i+1;
            }
        }

每个数都加一次，遇到大的就储存它的起点和终点。当和小于0时，统计和的那个变量初始化为0。

代码：

#include<iostream>
using namespace std;
int main()
{
    int T,i,d,p,start,end,k,t,max;
    int n;
    cin>>T;
    d=0;
    p=0;
    while(T--)
    {
        cin>>n;
        max=-9999;
        start=end=k=1;
        t=0;    
        for(i=1;i<=n;i++)
        {
            cin>>d;
            t+=d;
            if(t>max)
            {
                start=k;
                end=i;
                max=t;
            }
            if(t<0)
            {
                t=0;
                k=i+1;
            }
        }
        cout<<"Case "<<++p<<":"<<endl;
        cout<<max<<" "<<start<<" "<<end<<endl;
        if(T)
            cout<<endl;        
    }
    return 0;
}