zoukankan      html  css  js  c++  java
  • PAT Advanced 1007 Maximum Subsequence Sum (25 分)

    Given a sequence of K integers { N1​​, N2​​, ..., NK​​ }. A continuous subsequence is defined to be { Ni​​, Ni+1​​, ..., Nj​​ } where 1. 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 (≤). 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
    #include <iostream>
    #include <vector>
    using namespace std;
    
    int main()
    {
        /**计算时间
        * 开始7:14
        * 结束7:27
        */
        int T;
        cin>>T;int sum=0,tmp,max=-1;
        vector<int> vec;
        vector<int> res;
        vector<int> all;
        while(T--){
            cin>>tmp;
            all.push_back(tmp);
            vec.push_back(tmp);
            sum+=tmp;
            if(sum>max) {
                max=sum;
                res=vec;
            }
            if(sum<0){
                sum=0;
                vec.clear();
            }
        }
        if(max!=-1) cout<<max<<" "<<res[0]<<" "<<res[res.size()-1];
        else cout<<0<<" "<<all[0]<<" "<<all[all.size()-1];
        system("pause");
        return 0;
    }
  • 相关阅读:
    lvs+keepalived集群架构服务
    GlusterFS分布式存储系统
    Zabbix监控平台(三)生产环境案例
    Zabbix监控平台(二)深入了解
    Zabbix监控平台(一)搭建部署与概述
    Memcached数据库缓存
    Mariadb 基于Mycat实现读写分离
    基于mysqld_multi实现MySQL多实例配置
    tomcat-APR配置及三种工作模式简介
    Session服务器之Memcached与Redis
  • 原文地址:https://www.cnblogs.com/littlepage/p/11286389.html
Copyright © 2011-2022 走看看