ACM-DP之Max Sum——hdu1003

之前去忙齐鲁软件大赛了，好久没有更新CSDN了，

现在忙完了，继续ACM生活。。

今天做了一道动态规划的题目：

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 116583    Accepted Submission(s): 27023

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

//*******************************************************
//*  程 序 名：hdu_1003.cpp								*
//*														*
//*  作    者：Tree										*
//*														*
//*  编制时间：2013年9月27日							*
//*														*
//*  主要功能：hdu 1003 Max Sum							*
//*														*
//*******************************************************

/*
	思路：
		这是一道动态规划（DP）的题目，求所输入的一组数据中，连续的数组成的和最大，标出这段连续数的起始位置和终止位置，
		大体想法为：  读入一个数temp，之前数字之和为now，判断 now+temp
using namespace std;
int main()
{
	
	int now,temp,max,first,last;				//now为当前数据的和（不算正在读取的数据），temp为所读取的数据，max为最大数据和（所求值），
												//first为开始位置，last为终止位置
	int i,j,n,t,x;								//t为共多少组数据，n为每组数据个数，i为循环变量，j为Case输出准备,x临时记录起始位置
	
	
	cin>>t;

	for(j=1;j<=t;++j)
	{
		cin>>n>>temp;
		
		first=last=x=1;
		now=max=temp;
		

		for(i=2;i<=n;++i)
		{
			cin>>temp;
			if(now+tempmax)
			{
				max=now;
				first=x;
				last=i;
			}
		}

		
		cout<<"Case "<