HDU 1003 Max Sum(线性DP)

最新推荐文章于 2023-08-25 14:06:59 发布

原创最新推荐文章于 2023-08-25 14:06:59 发布 · 195 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#dp

线性DP 专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种求解给定整数序列中最大子序列和的算法，并提供了详细的实现思路与C语言代码示例。通过对每个元素为结尾的最大子序列和进行分析，实现了高效的求解过程。

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

思路：

以每个点为结尾求该点最大子序列和。

设i点权值为b，i-1点最大子序列和为a，如果a是负数那么以该点为结尾的最大子序列和为b，否则为a+b。

代码：

#include<stdio.h>
#define max(a,b) (a>b?a:b)

int main()
{
//l存储子序列和最大的左边界，r存储子序列和最大的右边界，c暂存当前序列左边界
//a存储前i-1项最大子序列和，b存储当前i项权值，maxn存储子序列最大和
    int T,n,i,a,b,maxn,cnt=0,l,r,c;
    scanf("%d",&T);
    while(T--)
    {
        
        scanf("%d%d",&n,&a);
        maxn=a,l=1,r=1,c=1;
        for(i=2;i<=n;i++)
        {
            scanf("%d",&b);
            if(a<0)c=i,a=b;
            else a+=b;

            if(a>maxn)
            {
                l=c;
                maxn=a;
                r=i;
            }
        }
        if(cnt++)printf("\n");
        printf("Case %d:\n%d %d %d\n",cnt,maxn,l,r);
    }
	return 0;
}