上升时间最长的序列

Problem Description

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

Author

HYNU

代码：

#include<stdio.h>
#include<malloc.h>
struct mm
{
   int a;// 输入的值 
   int b;// 存放长度 
}*x;
int main()
{
   int n,max,i,j;
   while(scanf("%d",&n)!=EOF)
   {
      x=(mm *)malloc(sizeof(mm)*n);
      for(i=0;i<n;i++)
        {                                  // 1 2 2 3 4 3 4   
           scanf("%d",&x[i].a);           // 1 7 3 5 9 4 8                                
           x[i].b=1;//每一个数相应长度为一 // 1 1 1 1 1 1 1
        }   
      for(i=0;i<n;i++)
        {
           for(max=0,j=0;j<i;j++)
             {
                if(x[i].a>x[j].a&&x[j].b>max)// 找当前数（i)比其前面的数大并相应的最长的长度
                  {
                     max=x[j].b;
                  }
             }
           x[i].b=max+1;// 前面的最长长度加上本身
        }
      for(i=0,max=0;i<n;i++)
         if(max<x[i].b)max=x[i].b;// 找出最长长度 
      printf("%d\n",max);  
      free(x);
   }
   return 0;
}

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