hdu-1003 Max sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 282173    Accepted Submission(s): 67021

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4

Case 2:7 1 6 。

最近在入门dp..

题目的意思是：找一段数的某个连续区间，这个连续区间的和最大，输出最大的和以及区间的始末位置

#include
#include
using namespace std;
int main()
{
    int n,t,m=1,a;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>t;
        int sum=0,start=0,endd=0,temp=1,maxn=-1005;
        for(int j=0;j>a;
            sum+=a;
            if(sum>maxn)
            {
                maxn=sum;
                start=temp;
                endd=j+1;
            }
            if(sum<0)//当sum<0;时要初始化sum
            {
                sum=0;
                temp=j+2;
            }
        }
        printf("Case %d:\n%d %d %d\n",m++,maxn,start,endd);
        if(i!=n)
            printf("\n");
    }
    return 0;
}