HDU 1003 Max Sum【最大子串和】

最新推荐文章于 2024-08-13 10:26:25 发布

原创最新推荐文章于 2024-08-13 10:26:25 发布 · 407 阅读

0 ·

CC 4.0 BY-SA版权

hdu oj 同时被 2 个专栏收录

269 篇文章

订阅专栏

最大子串和最大子矩阵

7 篇文章

订阅专栏

本文探讨了最大子串和问题的解决方案，通过一个具体示例介绍了如何计算给定序列的最大子串和，并记录该子串的起始与结束位置。文章提供了完整的C语言代码实现。

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): 195854 Accepted Submission(s): 45718

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

最大子串和，不过需要记录子串的开始和结束位置，那么就需要对这个动态更新的过程比较清楚，这里用了一个中转变量来记录开始位置，结束位置比较好找，注意三个操作的顺序！

#include<stdio.h>
int main()
{
	int t,n,x[100005];
//	freopen("shuju.txt","r",stdin);
	scanf("%d",&t);
	for(int k=0;k<t;++k)
	{
		if(k)
		{
			printf("\n");
		}
		scanf("%d",&n);
		for(int i=0;i<n;++i)
		{
			scanf("%d",&x[i]);
		}
		int ans=x[0],tp=0,l=0,r=0,m=0;
		for(int i=0;i<n;++i)
		{
			if(tp<0)
			{
				tp=0;m=i;
			}
			tp+=x[i];
			if(tp>ans)
			{
				ans=tp;l=m;r=i;
			}
		}
		printf("Case %d:\n%d %d %d\n",k+1,ans,l+1,r+1);
	}
	return 0;
}