Max Sum

先看看题目要求：

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26441    Accepted Submission(s): 5582

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

解题思路借鉴别人的，首先由题目可知，需要变量如下

t	        测试数据组数
n	        每组数据的长度
temp	        当前取的数据
pos1  	最后MAX SUM的起始位置
pos2	        最后MAX SUM的结束位置
max		当前得到的MAX SUM
now		在读入数据时，能够达到的最大和
x		记录最大和的起始位置，因为不知道跟之前的max值的大小比，所以先存起来

下面模拟过程：
1.首先，读取第一个数据，令now和max等于第一个数据，初始化pos1,pos2,x位置
2.然后，读入第二个数据，判断
①. 若是now+temp<temp，表示当前读入的数据比之前存储的加上当前的还大，说明可以在当前另外开始记录，更新now=temp
②. 反之，则表示之前的数据和在增大，更新now=now+temp
3.之后，把now跟max做比较，更新或者不更新max的值，记录起始、末了位置
4.循环2~3步骤，直至读取数据完毕。

#include <iostream>
using namespace std;

int main()
{
    int t,n,temp,pos1,pos2,max,now,x,i,j;
    cin>>t;
    for (i=1;i<=t;i++)
    {
        cin>>n>>temp;
        now=max=temp;
        pos1=pos2=x=1;
        for (j=2;j<=n;j++)
        {
            cin>>temp;
            if (now+temp<temp)
                now=temp,x=j;
            else
                now+=temp;
            if (now>max)
                max=now,pos1=x,pos2=j;
        }
        cout<<"Case "<<i<<":"<<endl<<max<<" "<<pos1<<" "<<pos2<<endl;
        if (i!=t)
            cout<<endl;
    }
    return 0;
}