Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 70077    Accepted Submission(s): 16106

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
  
  

 

Author

Ignatius.L

 

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1003

这个题目的话 主要是注意 全部都为负数 的情况 其他的就不多说了。。

#include<iostream>
using namespace std;
int main()
{
	long int T,N,a[100000],b,c,sum=-100000000,sun=0,i,j,t=1;
	cin>>T;
	while(T--)
	{
		cin>>N;
		for(i=0;i<N;i++)
			cin>>a[i];
		cout<<"Case "<<t<<':'<<endl;
		for(i=0;i<N;i++)
		{
			for(j=i;j<N;j++)
			{
				sun=sun+a[j];
				if(sun>sum)
				{
					sum=sun;
					b=i;
					c=j;
				}
			}
			sun=0;
		}
		cout<<sum<<' '<<b+1<<' '<<c+1<<endl;
		sum=-100000000;
		if(T>0)
			cout<<endl;
		t++;
	}
	return 0;
}