给定Q个查询，其中每个查询由整数N组成，任务是查找大于N的最小整数，以使其二进制表示形式中不存在连续的1 。
例子：
Input: Q[] = {4, 6}
Output:
5
8
Input: Q[] = {50, 23, 456}
Output:
64
32
512
方法：将所有数字存储在一个列表中，该列表的二进制表示形式不包含连续1到固定限制的数字。现在，对于每个给定的N ，在先前使用二进制搜索生成的列表中找到下一个更大的元素。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
const int MAX = 100000;
// To store the pre-computed integers
vector v;
// Function that returns true if the
// binary representation of x contains
// consecutive 1s
int consecutiveOnes(int x)
{
// To store the previous bit
int p = 0;
while (x > 0) {
// Check whether the previous bit
// and the current bit are both 1
if (x % 2 == 1 and p == 1)
return true;
// Update previous bit
p = x % 2;
// Go to the next bit
x /= 2;
}
return false;
}
// Function to pre-compute the
// valid numbers from 0 to MAX
void preCompute()
{
// Store all the numbers which do
// not have consecutive 1s
for (int i = 0; i <= MAX; i++) {
if (!consecutiveOnes(i))
v.push_back(i);
}
}
// Function to return the minimum
// number greater than n which does
// not contain consecutive 1s
int nextValid(int n)
{
// Search for the next greater element
// with no consecutive 1s
int it = upper_bound(v.begin(),
v.end(), n)
- v.begin();
int val = v[it];
return val;
}
// Function to perform the queries
void performQueries(int queries[], int q)
{
for (int i = 0; i < q; i++)
cout << nextValid(queries[i]) << "\n";
}
// Driver code
int main()
{
int queries[] = { 4, 6 };
int q = sizeof(queries) / sizeof(int);
// Pre-compute the numbers
preCompute();
// Perform the queries
performQueries(queries, q);
return 0;
} Java
// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG{
static int MAX = 100000;
// To store the pre-computed integers
static ArrayList v = new ArrayList();
public static int upper_bound(ArrayList ar,
int k)
{
int s = 0;
int e = ar.size();
while (s != e)
{
int mid = s + e >> 1;
if (ar.get(mid) <= k)
{
s = mid + 1;
}
else
{
e = mid;
}
}
if (s == ar.size())
{
return -1;
}
return s;
}
// Function that returns true if the
// binary representation of x contains
// consecutive 1s
static int consecutiveOnes(int x)
{
// To store the previous bit
int p = 0;
while (x > 0)
{
// Check whether the previous bit
// and the current bit are both 1
if (x % 2 == 1 && p == 1)
{
return 1;
}
// Update previous bit
p = x % 2;
// Go to the next bit
x /= 2;
}
return 0;
}
// Function to pre-compute the
// valid numbers from 0 to MAX
static void preCompute()
{
// Store all the numbers which do
// not have consecutive 1s
for(int i = 0; i <= MAX; i++)
{
if (consecutiveOnes(i) == 0)
{
v.add(i);
}
}
}
// Function to return the minimum
// number greater than n which does
// not contain consecutive 1s
static int nextValid(int n)
{
// Search for the next greater element
// with no consecutive 1s
int it = upper_bound(v,n);
int val = v.get(it);
return val;
}
// Function to perform the queries
static void performQueries(int queries[], int q)
{
for(int i = 0; i < q; i++)
{
System.out.println(nextValid(queries[i]));
}
}
// Driver code
public static void main(String[] args)
{
int queries[] = { 4, 6 };
int q = queries.length;
// Pre-compute the numbers
preCompute();
// Perform the queries
performQueries(queries, q);
}
}
// This code is contributed by rag2127 Python3
# Python3 implementation of the approach
from bisect import bisect_right as upper_bound
MAX = 100000
# To store the pre-computed integers
v = []
# Function that returns true if the
# binary representation of x contains
# consecutive 1s
def consecutiveOnes(x):
# To store the previous bit
p = 0
while (x > 0):
# Check whether the previous bit
# and the current bit are both 1
if (x % 2 == 1 and p == 1):
return True
# Update previous bit
p = x % 2
# Go to the next bit
x //= 2
return False
# Function to pre-compute the
# valid numbers from 0 to MAX
def preCompute():
# Store all the numbers which do
# not have consecutive 1s
for i in range(MAX + 1):
if (consecutiveOnes(i) == 0):
v.append(i)
# Function to return the minimum
# number greater than n which does
# not contain consecutive 1s
def nextValid(n):
# Search for the next greater element
# with no consecutive 1s
it = upper_bound(v, n)
val = v[it]
return val
# Function to perform the queries
def performQueries(queries, q):
for i in range(q):
print(nextValid(queries[i]))
# Driver code
queries = [4, 6]
q = len(queries)
# Pre-compute the numbers
preCompute()
# Perform the queries
performQueries(queries, q)
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
static int MAX = 100000;
// To store the pre-computed integers
static List v = new List();
static int upper_bound(List ar, int k)
{
int s = 0;
int e = ar.Count;
while (s != e)
{
int mid = s + e >> 1;
if (ar[mid] <= k)
{
s = mid + 1;
}
else
{
e = mid;
}
}
if (s == ar.Count)
{
return -1;
}
return s;
}
// Function that returns true if the
// binary representation of x contains
// consecutive 1s
static int consecutiveOnes(int x)
{
// To store the previous bit
int p = 0;
while (x > 0)
{
// Check whether the previous bit
// and the current bit are both 1
if (x % 2 == 1 && p == 1)
{
return 1;
}
// Update previous bit
p = x % 2;
// Go to the next bit
x /= 2;
}
return 0;
}
// Function to pre-compute the
// valid numbers from 0 to MAX
static void preCompute()
{
// Store all the numbers which do
// not have consecutive 1s
for(int i = 0; i <= MAX; i++)
{
if (consecutiveOnes(i) == 0)
{
v.Add(i);
}
}
}
// Function to return the minimum
// number greater than n which does
// not contain consecutive 1s
static int nextValid(int n)
{
// Search for the next greater element
// with no consecutive 1s
int it = upper_bound(v, n);
int val = v[it];
return val;
}
// Function to perform the queries
static void performQueries(int[] queries, int q)
{
for(int i = 0; i < q; i++)
{
Console.WriteLine(nextValid(queries[i]));
}
}
// Driver code
static public void Main()
{
int[] queries = { 4, 6 };
int q = queries.Length;
// Pre-compute the numbers
preCompute();
// Perform the queries
performQueries(queries, q);
}
}
// This code is contributed by avanitrachhadiya2155 输出：
5
8