Nyoj 214 Monotone Increment subsequence (ii)

Monotone increment subsequence (ii) time limit:MS | Memory limit:65535 KB Difficulty:4

Describe

Given an integer sequence {a1, a2..., an} (0<n<=100000), find out the monotone increment of the oldest eldest and calculate its length.

For example: 1 9 10 5 11 2 13 The longest monotonically incrementing subsequence is 1 9 10 11 13 with a length of 5.

Input

There are multiple sets of test data (<=7)
The first line of each set of test data is an integer n indicating that the sequence has n integers, followed by n integers in the next row, representing all the elements in the sequence. The middle of each shape number is separated by a space (0<n<=100000).
The data ends with EOF.
The input data is guaranteed to be legal (all int integers)!

Output

The length of the longest increment subsequence for each set of test data output shaping sequences, one row per output.

Sample input

71 9 10 5 11 2 1322-1

Sample output

51

1#include <iostream>2#include <cstdio>3 Const intMAX =100000+Ten;4 using namespacestd;5 intA[max], Dp[max];6 7 intBinary_search (intDigitintlength)8 {9     intleft=1, right =Length,mid;TenMid= (Right+left)/2; One      while(left<=Right ) A     { -         if(digit==Dp[mid]) -             returnmid; the         if(digit>Dp[mid]) -Left=mid+1; -         Else -right=mid-1; +Mid= (Right+left)/2; -     } +     returnLeft ; A } at intMain () - { -     intn,i,j,k; -      while(~SCANF ("%d", &N)) -     { -          for(i=0; i<n;i++) inscanf"%d", &a[i]); -         intlen=1; todp[1]=a[0]; +          for(i=1; i<n;i++) -         { thej=Binary_search (A[i],len); *dp[j]=A[i]; $             if(j>len)Panax Notoginsenglen=J; -         } theprintf"%d\n", Len); +     } A     return 0; the}

Nyoj 214 Monotone Increment subsequence (ii)