PAT 甲级 1007 Maximum Subsequence Sum(25)

Given a sequence of K integers { N​1​​, N​2​​, ..., N​K​​ }. A continuous subsequence is defined to be { N​i​​, N​i+1​​, ..., N​j​​ } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

Experiential Summing-up

 求最大连续子序列和。设置 sum 为最大和,temp 为临时最大和,l 和 r 分别是所求子序列的左右下标,t_l 为 l 的临时下标。

Accepted Code

复制代码
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int n;
 7     cin >> n;
 8     int v[n];
 9     int l = 0, r = n - 1, sum = -1, temp = 0, l_t = 0;
10     for (int i = 0; i < n; i++)
11     {
12         cin >> v[i];
13         temp = temp + v[i];
14         if (temp < 0)
15         {
16             temp = 0;
17             l_t = i + 1;
18         }
19         else if (temp > sum)
20         {
21             sum = temp;
22             l = l_t;
23             r = i;
24         }
25     }
26     if (sum < 0) sum = 0;
27     cout << sum << " " << v[l] << " " << v[r];
28     return 0;
29 }
复制代码

 

posted @   爱吃虾滑  阅读(14)  评论(0编辑  收藏  举报
相关博文:
阅读排行:
· 分享一个免费、快速、无限量使用的满血 DeepSeek R1 模型,支持深度思考和联网搜索!
· 基于 Docker 搭建 FRP 内网穿透开源项目(很简单哒)
· ollama系列01:轻松3步本地部署deepseek,普通电脑可用
· 25岁的心里话
· 按钮权限的设计及实现
点击右上角即可分享
微信分享提示