acm-1003

最新推荐文章于 2021-11-06 13:04:49 发布

青木落

最新推荐文章于 2021-11-06 13:04:49 发布

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分类专栏：算法文章标签：算法 acm

本文链接：https://blog.youkuaiyun.com/qq_41200424/article/details/80747006

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本文介绍了一种解决最大子序列和问题的有效方法，并通过一个示例程序详细展示了如何使用动态规划来找到给定序列中最大子序列的和及其位置。文章强调了正确处理输入输出及细节的重要性。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 286615 Accepted Submission(s): 68051

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

参考求最大子序列，只要处理一下输入输出以及一些细节即可，竞赛题细节很重要，培养了参赛选手的细心程度。

#include <iostream> //采用动态规划，用数组也是可以的，但容易超时 using namespace std; int main() { int T, N, sum, max, a, i, j, left, z, right; cin >> T; for(i = 0; i < T; i++) { cin >> N; for(left = z = right = sum = 0, max = -1001, j = 0; j < N; j++) { cin >> a; sum += a; if(max < sum) { left = z; right = j; max = sum; } if(sum < 0) { z = j + 1; sum = 0;//当总和小于0时就抛弃前面的值 } } cout << "Case " << i + 1 << ":" << endl; cout <<max << " " << ++left << " " << ++right << endl; if(i < T - 1) cout << endl; } return 0; }