hdu1003 Max Sum求最大和区间

最新推荐文章于 2023-01-28 16:57:25 发布

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最新推荐文章于 2023-01-28 16:57:25 发布

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分类专栏： 2.练手题目文章标签： hdu1003 java 最大和区间

本文链接：https://blog.youkuaiyun.com/qq_36738482/article/details/78650417

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本文介绍了一种求解子序列最大和的经典算法，并通过一个具体示例详细展示了其实现过程。该算法能够高效地找出给定整数序列中具有最大和的连续子序列及其起始和结束位置。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 263148 Accepted Submission(s): 62539

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

思路：

这里我直接用sum累加数组的每一个，如果sum>max那么终点和max就要改变，如果sum<0那么z改变，需要注意的是，这里不是改变起点，也就是star,只有当sum>max时，才用z改变Star的值。

import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner cin= new Scanner(System.in);
		int maxsum,maxhere,t;
		int a[]=new int[100000];
		int n,L,R,l,r;
		t=cin.nextInt();
		int count=0;
		//System.out.println(t);
		for(int j=0;j<t;j++)
		{
			count++;
			n=cin.nextInt();
			int max=-9999,sum=0;
			int star,end,z;
			L=R=star=end=z=0;
			for(int i=0;i<n;i++)
			{
				a[i]=cin.nextInt();
	                sum=sum+a[i];
	                if(max<sum){
	                    max=sum;
	                    end=i;
	                    star=z;
	                }
	                if(sum<0){
	                    sum=0;
	                    z=i+1;
	                }
			}
			System.out.println("Case "+count+":");
			System.out.println(max+" "+(star+1)+" "+(1+end));
			if(j!=t-1)
			System.out.println();
			//a=null;
		}
		cin.close();//关闭输入流，关闭所有。
	}
}