zoukankan      html  css  js  c++  java
  • Maximum Subsequence Sum

    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

    题目描述略坑

     1 #include<iostream>
     2 #include<cstdio>
     3 using namespace std;
     4 
     5 const int MAXN=10005;
     6 const int INF=0x7f7f7f7f;
     7 int k,x,y,maxn=-INF;
     8 int a[MAXN],f[MAXN];
     9 
    10 int main()
    11 {
    12     scanf("%d",&k);
    13     for(int i=1;i<=k;i++)
    14         scanf("%d",&a[i]);
    15     f[0]=-1;
    16     for(int i=1;i<=k;i++)
    17     {
    18         f[i]=max(f[i-1]+a[i],a[i]);
    19         if(f[i]>maxn)
    20             y=i,maxn=f[i];
    21     }
    22     for(x=y;x>=1;x--)
    23         if(f[x-1]<0) break;
    24     if(maxn<0)
    25         printf("0 %d %d",a[1],a[k]);
    26     else
    27         printf("%d %d %d",maxn,a[x],a[y]);
    28     return 0;
    29 }
  • 相关阅读:
    搜索引擎 中 排序学习 的小思考
    《算法导论》之分治策略与动态规划
    《算法导论》之基础篇
    中文文本信息处理的原理与应用读书笔记1
    python 类变量 在多线程下的共享与释放问题
    日志管理
    《领导梯队》读书分享
    初见微服务之服务注册与发现
    初见微服务之RESTful API
    初见微服务之架构概述
  • 原文地址:https://www.cnblogs.com/InWILL/p/10524736.html
Copyright © 2011-2022 走看看