Hdu 5439 Aggregated Counting)

Hdu 5439 Aggregated Counting)
Aggregated CountingTime Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission (s): 665 Accepted Submission (s): 302

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 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 ,....

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 wowould 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 the last position of each query N . In case the answer is greater 1000000006 , Please modulo the answer 1000000007 .
Sample Input

3 3 10 100000
Sample Output

11 217 507231491
Source 2015 ACM/ICPC Asia Regional Changchun Online
Recommend hujie | We have carefully selected several similar problems for you: 56645663566256615660 question: generate a sequence according to the following rules to make f (n) returns the value of f (n) for the last position of n in the sequence.

Solution:

To column a in the original order, the following rules can be introduced based on the definition and sequence generation rules:

• F (n) is equal to the first n of a and
• F (n) is the last position where n appears in.

  F (n) indicates the first m of a, and m indicates the position where n finally appears in. Therefore, the formula of f (n) can be written as follows:

  F (n) = 1 + (2 + 3) * 2 + (4 + 5) * 3 + (6 + 7 + 8) * 4 +... + (... + n) * t

  This is probably the above content. Here I use several arrays to access the above process and sort out the formula.

  For details, see the code.

   

  # Include
     
      
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  # Include using namespace std; # define N 450010 const int Mod = (1e9 + 7); long a [N]; // It indicates the ending number of I, int B [N]; // It indicates that I appears several times long s [N]; // indicates the first I and int main () {a [1] = 1; a [2] = 3; a [3] = 5; B [1] = 1; B [2] = 2; B [3] = 2; s [1] = 1; s [2] = 11; s [3] = 38; int k = 4, pre = 4, kk = 3; // k is used for loop, pre indicates the subscript of the current column, kk represents the number corresponding to the subscript while (a [k-1] <= 1e9) {for (k = pre; k
        
  a[mid-1])            {                break;            }            else            {                if (n>a[mid])                {                    l=mid+1;                }                else                    r=mid;            }        }        printf ("%lld\n",(s[mid-1]+mid*((a[mid-1]+1+n)*(n-a[mid-1])/2))%Mod);    }    return 0;}