题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

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

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

Author

Ignatius.L

题意: 求最大字段和，并给出字段的起始点和终点。

题解:DP之，模板题——O(n)的算法。

AC代码:

#include<iostream>
#define N 100000+5
using namespace std;
int dp[N],t,n;
int main()
{
    cin.sync_with_stdio(false);
    cin>>t;
    for(int k=1;k<=t;k++){
        cin>>n;
        for(int i=0;i<n;i++)cin>>dp[i];
        int sum=dp[0],tmp=dp[0],s=1,b=1,d=1;
        for(int i=1;i<n;i++){
            if(tmp>=0)
                tmp+=dp[i];
            else
                tmp=dp[i],s=i+1;
            if(tmp>sum) sum=tmp,d=i+1,b=s;
        }
        cout<<"Case "<<k<<":"<<endl;
        cout<<sum<<" "<<b<<" "<<d<<endl;
        if(k!=t)
            cout<<endl;
    }
    return 0;
}