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tc srm 632 500 (规律)

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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.

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

 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 };

 

tc srm 632 500 (规律)

标签:style   blog   color   os   io   ar   strong   for   div   

原文地址:http://www.cnblogs.com/bfshm/p/3958733.html

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