Aggregated Counting

Time Limit: 1500/1000 MS (Java/Others)Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 665Accepted 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, 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 numbern, 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 queryn. 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    
Source   2015 ACM/ICPC Asia Regional Changchun Online    
Recommend   hujie|We have carefully selected several similar problems for you:56645663566256615660     題目大意：：按下列規則生成一組序列，令f(n)為n這個數在序列中出現的最後一個位置，求f(f(n))的值。

解題思路：

令原序列為a，根據定義和序列的生成規則可以推出：

• f(n)等於a的前n項和
• f(n)是n這個數在a中出現的最後一個位置

  f(f(n))的含義為:a的前m項和，m為n在a中最後出現的位置。所以f(f(n))的計算式可以寫成：

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

  大概就是上述的內容，我這裡采用了幾個數組來存取上述的過程，把公式整理出來。

  詳見代碼。

   

  #include 
  #include 
  #include 
  #include 
  
  using namespace std;
  
  #define N 450010
  const int Mod=(1e9+7);
  
  long long a[N];//表示第i個的結尾數字
  int b[N];//表示i出現了幾次
  long long s[N];//表示前i個和
  
  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用來循環，pre表示的當前這一列數的下標，kk表示下標對應的數字是多少
      while (a[k-1]<=1e9)
      {
          for (k=pre;ka[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;
  }