hdu1003 Max Sum--DP-白红宇的个人博客

hdu1003 Max Sum--DP

发布日期：2021-10-03 20:31:50 浏览次数：6 分类：技术文章

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原题链接：

一：原题内容

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

二：分析理解

一个简单动态规划。。。。

三：AC代码

#include
   
    #include
    
     #include
     
      using namespace std;int a[100010];int main(){	int T;	scanf("%d", &T);	for (int i = 1; i <= T; i++)	{		scanf("%d", &a[0]);		for (int j = 1; j <= a[0]; j++)			scanf("%d", &a[j]);		int left = 1;		int right = 1;		int cnt = a[1];		int flag = 1;		int max = a[1];		for (int k = 2; k <= a[0]; k++)		{			if (cnt < 0)			{				flag = k;				cnt = a[k];			}			else				cnt += a[k];			if (cnt > max)			{				left = flag;				right = k;				max = cnt;			}		}		printf("Case %d:\n%d %d %d\n", i, max, left, right);		if (i != T)			printf("\n");	}	return 0;}

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