hduoj1003,Max Sum,最大子序列和，类动态规划，经典

Max Sum

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

分析：

最大子序列和问题

用两组变量startn,endn,startt,endt分别存储最终的起点终点和暂时的起点终点。

考虑所有元素都为负的情况，最大和即为最大的那个数。

当和有正数时，按次加上后一个元素与最大和比较。

code：

#include<iostream>
#define MAXSIZE 100000
using namespace std;
int a[MAXSIZE+20];
int main()
{
    int n,m,i,sum,msum,Case=0;
    cin>>n;
    while(n--)
    {
        int startn,endn,startt,endt;
        msum=sum=-9999;
        memset(a,0,sizeof(a));
        cin>>m;
        for(i=0;i<m;i++)
            cin>>a[i];
        for(i=0;i<m;i++)
        {
            if(sum<0)
            {
                if(a[i]>sum)
                {
                    sum=a[i];
                    startt=endt=i;
                    if(msum<sum)
                    {
                        msum=sum;
                        startn=startt;
                        endn=endt;
                    }
                }
            }
            else
            {
                sum+=a[i];
                endt=i;
                if(msum<sum)
                {
                    msum=sum;
                    startn=startt;
                    endn=endt;
                }
            }
        }
        Case++;
        cout<<"Case "<<Case<<":"<<endl;
        cout<<msum<<" "<<startn+1<<" "<<endn+1<<endl;
        if(n) cout<<endl;
    }
    return 0;
}