HDU1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 127665    Accepted Submission(s): 29582

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

 

刚开始我想的方法是把子数组全部找到，用sum_max，min,max存放结果超时了，然后看到这种解法
下面是AC代码：

#include<iostream>
#include<string>
#include<map>
using namespace std;
int a[100000];
int main()
{
	int n,m,i,j,flag=1;
	cin>>n;
	while(n--)
	{
		
		cin>>m;
		for(i=0;i<m;i++)
			cin>>a[i];
		cout<<"Case "<<flag<<':'<<endl;
		int mi=0,ma=0;//mi是开始，ma是结尾
		int pos=0;//pos作为一个存放临时初始位置
		int sum,max_sum;//sum存放变化的最大值，max_sum存放最终的最大值
	        sum=max_sum=a[0];
		for(i=1;i<m;i++)
		{ 				
			 sum+=a[i];
			if(a[i]>sum)
				{
					sum=a[i];
					pos=i;
				}//找到起始位置i
			if(sum>max_sum)
			{
				max_sum=sum;
				mi=pos;
				ma=i;
			}//不断的更新最大值，然后记下最后的位置
			
		}
		cout<<max_sum<<' '<<mi+1<<' '<<ma+1<<endl;
		flag++;
		if(n!=0)cout<<endl;//格式要求，最后一行没有空格
	}
	return 0;
}