E

E - 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        
71 7 3 5 9 4 8
            
Sample Output        
4

嗯就是最长升成子序列的问题 然后就是 主要的关键点在于要利用二分啊 心痛没学扎实没好意思说 

代码：

#include<cstdio>#include<cstring>#define MAXN 1005int arr[MAXN],ans[MAXN],len;int search(int k){int left=0;int right=len;while(left<right){int mid=(left+right)/2;if(ans[mid]>=arr[k]) right=mid;else left=mid+1;}return left;}//这个东西求得是最小(是最小哦 这样子才能替换)比它大的数的位置 嗯我觉得这一点理解很重要 int main(){int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&arr[i]);len=1;ans[1]=arr[1];for(int i=2;i<=n;i++){if(arr[i]>ans[len])ans[++len]=arr[i];else{int pos=search(i);ans[pos]=arr[i];}}printf("%d\n",len);return 0;}

然后还有就是要注意的就是比如

2

2 5 7 3

虽然输出的长度是3  但是其实吧   排在ans里面的东西是2 3 7   嗯 把5给替换了就是这个样子