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-8;
const int maxn = 100+10;
using namespace std;

class PotentialGeometricSequence
{
 public:
    int numberOfSubsequences(vector <int> d)
    {
        int n = d.size();
        int i, j, ret = n+n-1, f, x, k, y;
        for(i = 2; i < n; i++)
        {
            for(j = 0; j < n; j++)
            {
                if(j+i < n)
                {
                    f = 0;
                    x = d[j+1]-d[j];
                    for(k = j+2; k <= j+i; k++)
                    {
                        y = d[k]-d[k-1];
                        if(y!=x)
                        f = 1;
                    }
                    if(f == 0)
                    ret ++;
                }
            }
        }
        return ret;
    }
};