zoukankan      html  css  js  c++  java
  • Longest Ordered Subsequence

    A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence ( a1, a2, ..., aN) be any sequence ( ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

    Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.
     
    Input
    The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000
     
    Output
    Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
     
    Sample Input
    7
    1 7 3 5 9 4 8
    Sample Output
    4
     1 //简单动态规划问题
     2 
     3 #include <stdio.h>
     4 
     5 #define MAX 1005
     6 
     7 int main()
     8 {
     9     int n;
    10     int num[MAX];
    11     int maxLen[MAX];
    12     int max = -1;
    13     int i,j;
    14     scanf("%d",&n);
    15     for (i=1; i<=n; i++)
    16       scanf("%d",&num[i]);
    17     for (i=1; i<=n; i++)
    18       maxLen[i] = 1;
    19     for (i=1; i<=n; i++)
    20       for (j=1; j<i; j++)
    21     {
    22       if(num[i]>num[j]&&maxLen[i]<=maxLen[j])
    23         maxLen[i] = maxLen[j]+1;
    24     }
    25     for (i=1; i<=n; i++)
    26       if (maxLen[i]>max)
    27         max = maxLen[i];
    28     printf("%d",max);
    29     return 0;
    30 }
    蒹葭苍苍,白露为霜; 所谓伊人,在水一方。
  • 相关阅读:
    Delphi 连接 Paradox
    编译MangosZero
    关于StartCoroutine的简单线程使用
    cocos2dc-x解决中文乱码
    C++类构造函数初始化列表
    dynamic_cast
    cocos2d-x for android:SimpleGame分析
    C++宏定义详解
    四 AndEngine 画线
    三 最简单的 AndEngine 程序框架
  • 原文地址:https://www.cnblogs.com/huwt/p/9980263.html
Copyright © 2011-2022 走看看