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 ( a1a2, ..., aN) be any sequence ( ai1ai2, ..., 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

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    using namespace std;
    #define N 10005
    int a[N], dp[N];
    
    int main()
    {
        int n, m, i, j, sum;
        while(scanf("%d", &n)!=EOF)
        {
            memset(a, 0, sizeof(a));
            memset(dp, 0, sizeof(dp));
            for(i=0;i<n;i++)
            {
                scanf("%d", &a[i]);
            }
            sum=0;
            for(i=1;i<n;i++)
            {
                for(j=0;j<i;j++)
                {
                    if(a[i]>a[j])
                    {
                        dp[i]=max(dp[j]+1, dp[i]);
                    }
                }
               sum=max(sum, dp[i]);
            }
            printf("%d\n", sum+1);
        }
        return 0;
    }

    还有一个错误的,因为过了不少样例,卡着我一直找不出来

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    using namespace std;
    #define N 10005
    int a[N], dp[N];
    
    int main()
    {
        int n, m, i, j, sum;
        while(scanf("%d", &n)!=EOF)
        {
            memset(a, 0, sizeof(a));
            memset(dp, 0, sizeof(dp));
            for(i=0;i<n;i++)
            {
                scanf("%d", &a[i]);
            }
            dp[0]=1;
            sum=0;
            for(i=1;i<n;i++)
            {
                for(j=0;j<i;j++)
                {
                    if(a[i]>a[j])
                    {
                        dp[i]=max(dp[j], dp[i]);
                    }
                }
                dp[i]++;
                sum=max(sum, dp[i]);
            }
            printf("%d\n", sum);
        }
        return 0;
    }
  • 相关阅读:
    不得不爱开源 Wijmo jQuery 插件集(6)【Popup】(附页面展示和源码)
    遗漏的知识点
    初识函数
    ==和is的区别 以及编码和解码
    函数的动态参数 及函数嵌套
    基本数据类型补充、set集合、深浅拷贝
    文件操作
    基本数据类型之“字典”
    建立自己的Servlet
    还原误删数据笔记
  • 原文地址:https://www.cnblogs.com/zct994861943/p/6800123.html
Copyright © 2011-2022 走看看