hdu 1003 Max sum(简单DP)

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

题意：求最大连续和，并输出该连续和的起始、结尾位置

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=100010;

int a,b,dp[maxn],pre[maxn];

int main()
{
    int m,n,T,cas;
    cin>>T;
    for(cas=1;cas<=T;cas++)
    {
        cout<<"Case "<<cas<<':'<<endl;
        cin>>n;
        cin>>dp[1];
        m=pre[1]=1;

        for(int i=2; i<=n; i++ )
        {
            cin>>a;
            if(dp[i-1]>=0)//注意这里是大于'等于'0;
            {
                dp[i]=a+dp[i-1];
                pre[i]=pre[i-1];
            }
            else
                dp[i]=a,pre[i]=i;
            if( dp[i] > dp[m] )
                    m=i;//m保存当前最大连续和的下标
        }

        cout<<dp[m]<<' '<<pre[m]<<' '<<m<<endl;
        if(cas!=T) cout<<endl;
    }
    return 0;
}