hdu 1003Max Sum（dp）

最新推荐文章于 2021-02-25 23:35:01 发布

原创最新推荐文章于 2021-02-25 23:35:01 发布 · 282 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #hdu

编程算法专栏收录该内容

75 篇文章

订阅专栏

本文详细介绍了MaxSum问题的解决方案，包括算法原理、输入输出规范及实例演示，旨在帮助读者理解如何高效地解决序列中子段的最大和问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Max Sum

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

Total Submission(s): 209514 Accepted Submission(s): 49113

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

注意格式最后一行没有空格

#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int main()
{
	long long n,m,i,k,begin,bbegin,end,eend,max,sum,t;
	cin>>k;
	n=1;
	while(k--)
	{
		cin>>m;
		for(i=1;i<=m;i++)
		{
			cin>>t;
			if(i==1)
			{
				sum=max=t;
				bbegin=begin=eend=end=1;
			}
			else {
				if(sum+t<t)
				{
					sum=t;
					begin=i;
				}
				else sum+=t;
			}
			if(sum>max)
			{
				max=sum;
				bbegin=begin;
				eend=i;
			}
		}
		cout<<"Case "<<n++<<":"<<endl;
		cout<<max<<" "<<bbegin<<" "<<eend<<endl;
		if(k)cout<<endl;
	}
	return 0;
}