最大子段和

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 120365    Accepted Submission(s): 27835

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

//动规思想
#include<iostream>
#define MIN -999999
using namespace std;

int main()
{
	int T,i,j,sum,n,cnt,start,end,a[100001],ans;

	cin>>T;

	cnt=0;
	while(cnt<T)
	{
		cin>>n;

		cnt++;
		ans = MIN;
		sum=0;
		j=1;//临时起始下标；

		for(i=1;i<=n;i++)
		{
			cin>>a[i];
			sum+=a[i];
			if(sum>ans){ans=sum;start=j;end=i;}
			if(sum<0){sum=0;j=i+1;}
		}
		cout<<"Case "<<cnt<<":"<<endl;
		cout<<ans<<" "<<start<<" "<<end<<endl;
		if(cnt!=T)cout<<endl;
	}
	system("pause");
	return 0;
}