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. |
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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
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;
}
};
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