HDU 1003 (最大子序列和)

最新推荐文章于 2021-07-27 21:50:39 发布

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最新推荐文章于 2021-07-27 21:50:39 发布

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分类专栏： HDU

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本文介绍了一种求解整数序列中具有最大和的连续子序列的问题，并提供了一个具体的算法实现示例。输入包括多组测试数据，每组包含一系列整数；输出则为每组测试数据中找到的最大子序列和及其起始与结束位置。

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Max Sum

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
#include <stdio.h>
int main()
{
	int t,j;
	scanf("%d",&t);
    for(j=1;j<=t;j++)
	{
		int n,i;
		int sum=0,max=-2000,start=1,temp=1,end;
		scanf("%d",&n);
		int a[n+1];
		for(i=1;i<=n;i++)
		{
		   scanf("%d",&a[i]);			  
			   sum+=a[i];
			 if(max<=sum)
			   {
			   	start=temp;
			     max=sum;
			     end=i;
			   }
		    if(sum<0)
			{
			   sum=0;
			   temp=i+1;
		    }
		}
		printf("Case %d:\n%d %d %d\n",j,max,start,end);
		if(j!=t)
		printf("\n");
	}
}