最大子段和

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87125#problem/A

题目：

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：

    求出和最大的子数组，输出和，子数组的第一个和最后一个在原数组中的位置。

分析：

    直接一边遍历一边比较即可（记得更新子数组的起始位置和末尾位置）

 

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=100005;
int a[maxn],sum;
int main()
{
   int t,c=0,Max;
    cin>>t;
   while(t--)
   {
       c++;
       Max=-10000;
       sum=0;
     int n,k=1,i,j=0,x=0;
      cin>>n;
     for(i=1;i<=n;i++)
         cin>>a[i];
     for(i=1;i<=n;i++)
     {
      sum=sum+a[i];
     if(sum>Max)
     {     
         Max=sum;
        j=i;
        x=k;
     }
     if(sum<0)
     {
     sum=0;
     k=i+1;
     }
     }
     printf("Case %d:
",c);
     printf("%d %d %d
",Max,x,j);
     if(t)  cout<<endl;
   }
   return 0;
}