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[100005];
int a[100005];
int b[100005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,i,j;
        scanf("%d",&n);
        for(i = 1 ; i <= n  ; i++)
        {
            scanf("%d",&a[i]);
            dp[i]=0;
            b[i]=INF;
        }
        for(i = 1 ; i <= n ; i++)
        {
            int k=lower_bound(b+1,b+n+1,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 = 1 ; i < n ; i++)
            printf("%d ",dp[i]);
        printf("%d
",dp[n]);
    }
 }