杭电OJ 1003 Max Sum

 

                                           Max Sum

                   Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                      Total Submission(s): 315645    Accepted Submission(s): 75089

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

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DP算法

#include<iostream>
using namespace std;
void getMaxSum(int ncase); 
int main()
{
	int nCase,i;
	while(scanf("%d",&nCase)!=EOF)
	{
		for(i=1;i<=nCase;i++)
		{
			getMaxSum(i);
			if(i!=nCase)
			{
				cout<<endl;
			}
    	}
	}
	return 0;
} 
void getMaxSum(int ncase)
{
 	int from,to,sum,num;
 	int maxSum,maxFrom;
 	from=1;
 	to=1;
 	maxSum=-10001;
 	maxFrom=1;
 	sum=0;
 	cin>>num;
	int temp;
 	for(int i=1;i<=num;i++)
 	{
 		cin>>temp;
 		sum+=temp;
 		if(maxSum<sum)
		{
			maxSum=sum;
			maxFrom=from;//确定当前最大字段和的from 
			to=i;	
		}
 		if(sum<0)
 		{
 			sum=0;
 			from=i+1;//更新from 
		}
	}
	cout<<"Case "<<ncase<<":"<<endl;
	cout<<maxSum<<" "<<maxFrom<<" "<<to<<endl;
}