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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. |
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Definition |
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Limits |
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Notes |
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| - | 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 |
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| - | 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
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cstdlib>
5 #include <cmath>
6 #include <algorithm>
7 #include <vector>
8 #define LL __int64
9 const double eps = 1e-8;
10 const int maxn = 100+10;
11 using namespace std;
12
13 class PotentialGeometricSequence
14 {
15 public:
16 int numberOfSubsequences(vector <int> d)
17 {
18 int n = d.size();
19 int i, j, ret = n+n-1, f, x, k, y;
20 for(i = 2; i < n; i++)
21 {
22 for(j = 0; j < n; j++)
23 {
24 if(j+i < n)
25 {
26 f = 0;
27 x = d[j+1]-d[j];
28 for(k = j+2; k <= j+i; k++)
29 {
30 y = d[k]-d[k-1];
31 if(y!=x)
32 f = 1;
33 }
34 if(f == 0)
35 ret ++;
36 }
37 }
38 }
39 return ret;
40 }
41 };