zoukankan      html  css  js  c++  java
  • 浙大数据结构课后习题 练习一 7-1 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()
    {
        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:
    Codeforces Round #508 (Div. 2) C D
    Codeforces Round #493 (Div. 2)
    ACM-ICPC 2015 ChangChun
    ACM-ICPC 2015 BeiJing
    CodeFroces-- 514.div2.C-Sequence Transformation
    [Windows Server 2012] 网页Gzip压缩
    [Windows Server 2008] 安装网站伪静态
    [Windows Server 2003] 安装SQL Server 2005
    [Windows Server 2003] 安装PHP+MySQL方法
    [Windows Server 2003] IIS自带FTP安装及配置方法
  • 原文地址:https://www.cnblogs.com/littlepage/p/11374981.html
Copyright © 2011-2022 走看看