杭电 5748 Bellovin

Description

Peter has a sequence  and he define a function on the sequence -- , where  is the length of the longest increasing subsequence ending with .

Peter would like to find another sequence  in such a manner that  equals to . Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one.

The sequence  is lexicographically smaller than sequence , if there is such number  from  to , that  for  and .

Input

There are multiple test cases. The first line of input contains an integer , indicating the number of test cases. For each test case:

The first contains an integer   -- the length of the sequence. The second line contains  integers  .

Output

For each test case, output  integers   denoting the lexicographically smallest sequence. 

Sample Input

3
1
10
5
5 4 3 2 1
3
1 3 5

Sample Output

1
1 1 1 1 1
1 2 3

英语题目很难度，但是题意很简单，给你一个整数序列，输出以每个数结尾的最长上升子序列的长度。

 #include<cstdio>
 #include<algorithm>
 #define INF 0x3f3f3f3f
 using namespace std;
 int dp[];
 int a[];
 int b[];
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         int n,i,j;
         scanf("%d",&n);
         for(i =  ; i <= n  ; i++)
         {
             scanf("%d",&a[i]);
             dp[i]=;
             b[i]=INF;
         }
         for(i =  ; i <= n ; i++)
         {
             int k=lower_bound(b+,b+n+,a[i])-b;        //函数返回值为大于a[i]的第一个值的位置，比如序列1 1 4 5，a[i]为2，返回值为3
             dp[i]=k;                                    //与二分法类似，将数放入新数组
             b[k]=a[i];
         }
         for(i =  ; i < n ; i++)
             printf("%d ",dp[i]);
         printf("%d\n",dp[n]);
     }
  }