2015长春网络赛1003（hdu5439）推公式

Aggregated Counting

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 50    Accepted Submission(s): 19

Problem Description

Aggregated Counting Meetup (ACM) is a regular event hosted by Intercontinental Crazily Passionate Counters (ICPC). The ICPC people recently proposed an interesting sequence at ACM2016 and encountered a problem needed to be solved.

The sequence is generated by the following scheme.
1. First, write down 1, 2 on a paper.
2. The 2nd number is 2, write down 2 2’s (including the one originally on the paper). The paper thus has 1, 2, 2 written on it.
3. The 3rd number is 2, write down 2 3’s. 1, 2, 2, 3, 3 is now shown on the paper.
4. The 4th number is 3, write down 3 4’s. 1, 2, 2, 3, 3, 4, 4, 4 is now shown on the paper.
5. The procedure continues indefinitely as you can imagine. 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, . . . .

The ICPC is widely renowned for its counting ability. At ACM2016, they came up with all sorts of intriguing problems in regard to this sequence, and here is one: Given a positive number  n , First of all, find out the position of the last  n  that appeared in the sequence. For instance, the position of the last 3 is 5, the position of the last 4 is 8. After obtaining the position, do the same again: Find out the position of the last (position number). For instance, the position of the last 3 is 5, and the position of the last 5 is 11. ICPC would like you to help them tackle such problems efficiently.

 

Input

The first line contains a positive integer  T,T≤2000 , indicating the number of queries to follow. Each of the following  T  lines contain a positive number  n(n≤109)  representing a query.

 

Output

Output the last position of the last position of each query  n . In case the answer is greater than  1000000006 , please modulo the answer with  1000000007 .

 

Sample Input

   
   
   
   

    
    
    
    3
3
10
100000
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    11
217
507231491
   
   
   
   

 

题解：

 可以令n=1+2+2+3+3+......+ i    这个序列的长度为p

那么a[n]=1*1+2*2+3*2+...... + p*i 

那么不难发现a[a[n]] = 1*1 + (2+3)*2 + (4+5)*3 + (6+7+8)*4 + ... + (pre+1 + pre+2 + ... + pre+b[p] ) * p

b[p]为p在原序列中出现的次数

pre，b[p]这些值都可以预处理算出   每次询问可以O（1）算出答案

pre 是 第p-1项的pre + b[p-1]

                比赛没时间想了哎，赛后一推就出来了，果然还是太弱- -！