Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 174588 Accepted Submission(s): 40639

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

题意：给定一个序列，求其子序列的最大和，还有最大和的子序列的起始位置和结束位置。

求和与end都没什么好说的。求start是亮点，一开始判断了很久，后来还是那个想法，对于序列或是字符串正着想想不出来就逆过来想，从左到右end容易求，那从右至左的话start就容易求了。

代码：

#include <iostream>

#include <string>

#include <cstring>

#include <algorithm>

#include <cmath>

#pragma warning(disable:4996)

using namespace std;

int value[100005];

int dp[100005];

int main()

{

	//freopen("input.txt","r",stdin);

	//freopen("out.txt","w",stdout);

	int Test,num,i,j,ans,start,end;

	value[0]=0;

	cin>>Test;

	for(j=1;j<=Test;j++)

	{

		ans = -1005;

		memset(dp,0,sizeof(dp));

		cin>>num;

		for(i=1;i<=num;i++)

		{

			cin>>value[i];

			dp[i]=max(dp[i-1]+value[i],value[i]);

			if(dp[i]>ans)

			{

				ans=dp[i];

				end=i;

			}

		}

		ans=-1005;

		memset(dp,0,sizeof(dp));

		for(i=num;i>=1;i--)

		{

			dp[i]=max(dp[i+1]+value[i],value[i]);

			if(dp[i]>=ans)

			{

				ans=dp[i];

				start=i;

			}

		}

		cout<<"Case "<<j<<":"<<endl;

		cout<<ans<<" "<<start<<" "<<end<<endl;

		if(j!=Test)

			cout<<endl;

	}

	return 0;

}

后来看discuss里其他人的思路，觉得从end开始往回倒，对每一个值都加起来，什么时候等于求出来的最大和了，什么时候就是start了，觉得这样做也很好。

巴特西

HDU 1003：Max Sum

Max Sum

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