【DP经典系列】最大连续子序列和

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

题意：仍旧是最大连续子序列和；

注意输出格式就行；

AC代码：

#include 
#define ll long long
using namespace std ;
int main()
{
    int l , r , maxi , temp , now  ;
    int p1,  p2 ;
    int t ;
    cin>> t ;
    for(int cas = 1 ; cas <=t ; cas++)
    {
        int n ;
        cin>>n>>temp ;
        now=maxi=temp;
        r=l=p1=p2=1;
        for(int i=2; i<=n; i++)
        {
            cin>>temp;
            if(temp>now+temp)
            {
                now=temp;
                p1=i;
            }
            else
            {
                now+=temp;
            }
            if(now>maxi)
            {
                l=p1;
                r=i;
                maxi=now;
            }

        }
        printf("Case %d:\n",cas);
        printf("%d %d %d\n",maxi , l , r);
        if(cas!=t)
            printf("\n");
    }
    return 0 ;
}