HDU 1003-Max Sum

本文介绍了一种求解整数序列中最大和子序列的算法,并提供了完整的C++实现代码。通过动态规划的方法,该算法能高效地找出序列中具有最大和的连续子序列及其位置。

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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
 

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#include <iostream>
using namespace std;
int main()
{
	int a[100005],left,right,n,cnt=1,m,temp;
	cin>>n;
	while(n--)
	{
	
		left=right=temp=1;
		int maxx=-1000;
		int sum=0;
		cin>>m;
		for(int i=1;i<=m;i++)
		{
			cin>>a[i];
			sum+=a[i];
			if(sum>maxx)
			{
				maxx=sum;
				left=temp;
				right=i;	
			}
			if(sum<0)
			{
				sum=0;
				temp=i+1;
			}	
		}
		printf("Case %d:\n",cnt++);
		printf("%d %d %d\n",maxx,left,right);
		if(n>0) printf("\n");
	}return 0;
}

right 记录 最大值时 的 终点下标(终点)

left是起点 

如果和小于0,则从下一位重置起始值与和。

刚刚接触 动态规划。

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
 

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### HDU 4190 编程问题解析 针对HDU-4190这一特定编程挑战,该题目属于动态规划(DP)类问题[^3]。这类问题通常涉及寻找最优路径或者计算最优化的结果,在给定约束条件下实现目标最大化或最小化。 对于此题目的具体描述提到的是一个数塔结构,其中要求从顶部到底部移动,并且每次只能前往相邻节点,最终目的是使得所经过节点数值总和达到最大值。解决此类问题的关键在于理解如何有效地利用已知条件来构建解决方案: #### 动态规划算法设计 为了高效求解这个问题,可以采用自底向上的方法来进行动态规划处理。通过定义状态转移方程,逐步累积中间结果直至获得全局最优解。 ```python def max_sum_path(triangle): n = len(triangle) # 初始化dp数组用于存储各层的最大累加和 dp = [[0]*i for i in range(1, n+1)] # 设置起点即三角形顶端元素作为初始值 dp[0][0] = triangle[0][0] # 填充dp表 for level in range(1, n): for pos in range(level + 1): if pos == 0: dp[level][pos] = dp[level - 1][pos] + triangle[level][pos] elif pos == level: dp[level][pos] = dp[level - 1][pos - 1] + triangle[level][pos] else: dp[level][pos] = max(dp[level - 1][pos], dp[level - 1][pos - 1]) + triangle[level][pos] return max(dp[-1]) triangle = [ [2], [3, 4], [6, 5, 7], [4, 1, 8, 3] ] print(max_sum_path(triangle)) ``` 上述代码实现了基于输入参数`triangle`(表示数塔的数据结构)的函数`max_sum_path()`,它返回从顶至底所能得到的最大路径和。这里采用了二维列表形式保存每一级的最佳选择情况,从而保证能够快速访问并更新所需的信息。
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