杭电1003 Max Sum

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)
Total Submission(s): 49213    Accepted Submission(s): 10950

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

#include<iostream>
#include<vector>
using namespace std;

#define P(x) \
cout << #x " " << x << ":" << endl;

#define PE(max, beg, end) \
cout << max << " " << beg+1 << " " << end+1 << endl;

int max_sum(int &beg, int &end, vector<int> ivec)
{
    while(ivec[beg] < 0 && beg < ivec.size())
    {
        //忽略首相的负数
        beg++;
    }
    int max =  0, now = 0;
    for(int i = beg; i < ivec.size(); ++i)
    {
        now += ivec[i];
        if(max < now)
        {
            max =  now;
            end = i;
        }
    }
    return max;
}
int main()
{
    int T, Case = 1;
    cin >> T;
    while(T--)
    {
        vector<int> ivec;
        int n, x;
        cin >> n;
        while(n--)
        {
            cin >> x;
            ivec.push_back(x);
        }
        int beg = 0, end = 0, now_beg = 0, now_end = 0, now = -1000, max = -1000;
        for(; now_beg < ivec.size(); ++now_beg)
        {
            now = max_sum(now_beg, now_end, ivec);
            if(max < now)
            {
                max = now;
                beg = now_beg;
                end = now_end;
            }
        }
        P(Case);
        Case++;
        PE(max, beg, end);
        if(T)
            cout << endl;
    }
    return 0;
}

#include <iostream>
using namespace std;
int main()
{
    int T,N,num,startP,endP;
    cin>>T;
    for(int k=0;k<T;k++)
    {
        cin>>N;
        int max=-1001,sum=0,temp=1;
        for(int i=0;i<N;i++)
        {
            cin>>num;
            sum+=num;
            if(sum>max)
            {
                max=sum;
                startP=temp;
                endP=i+1;
            }
            if(sum<0)
            {
                sum=0;
                temp=i+2;
            }
        }
        cout<<"Case "<<k+1<<":"<<endl<<max<<" "<<startP<<" "<<endP<<endl;
        if(k!=T-1) cout<<endl;
    }
    return 0;
}