tc srm 632 500 (规律)

We have a sequence of N positive integers: a[0] through a[N-1]. You do not know these integers. All you know is the number of trailing zeros in their binary representations. You are given a vector <int> d with N elements. For each i, d[i] is the number of trailing zeros in the binary representation of a[i]. For example, suppose that a[0]=40. In binary, 40 is 101000 which ends in three zeros. Therefore, d[0] will be 3. You like geometric sequences. (See the Notes section for a definition of a geometric sequence.) You would like to count all non-empty contiguous subsequences of the sequence a[0], a[1], ..., a[N-1] that can be geometric sequences (given the information you have in d). More precisely: For each pair (i,j) such that 0 <= i <= j <= N-1, we ask the following question: "Given the values d[i] through d[j], is it possible that the values a[i] through a[j] form a geometric sequence?" For example, suppose that d = {0,1,2,3,2}. For i=0 and j=3 the answer is positive: it is possible that the values a[0] through a[3] are {1,2,4,8} which is a geometric sequence. For i=1 and j=4 the answer is negative: there is no geometric sequence with these numbers of trailing zeros in binary. Compute and return the number of contiguous subsequences of a[0], a[1], ..., a[N-1] that can be geometric sequences.
Definition
	Class: PotentialGeometricSequence Method: numberOfSubsequences Parameters: vector <int> Returns: int Method signature: int numberOfSubsequences(vector <int> d) (be sure your method is public)
Limits
	Time limit (s): 2.000 Memory limit (MB): 256
Notes
-	A geometric sequence is any sequence g[0], g[1], ..., g[k-1] such that there is a real number q (the quotient) with the property that for each valid i, g[i+1] = g[i]*q. For example, {1,2,4,8} is a geometric sequence with q=2, {7,7,7} is a geometric sequence with q=1, and {18,6,2} is a geometric sequence with q=1/3.
Constraints
-	N will be between 1 and 50, inclusive.
-	d will contain exactly N elements.
-	Each element of d will be between 0 and 100, inclusive.

d是二进制下这个数的末尾的0的个数，求其子序列里能够构成的等比序列的个数。

分析：求其等差子序列的个数

这应该算是一个看数字的规律题吧，我找的也挺慢的，不过想想二进制下每一位代表的数字 比较 一下规格，应该不难猜出来这个 等比 和 等差的规律

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #define LL __int64
 const double eps = 1e-;
 const int maxn = +;
 using namespace std;

 class PotentialGeometricSequence
 {
  public:
     int numberOfSubsequences(vector <int> d)
     {
         int n = d.size();
         int i, j, ret = n+n-, f, x, k, y;
         for(i = ; i < n; i++)
         {
             for(j = ; j < n; j++)
             {
                 if(j+i < n)
                 {
                     f = ;
                     x = d[j+]-d[j];
                     for(k = j+; k <= j+i; k++)
                     {
                         y = d[k]-d[k-];
                         if(y!=x)
                         f = ;
                     }
                     if(f == )
                     ret ++;
                 }
             }
         }
         return ret;
     }
 };