给定n，找到严格地不大于n且其二进制表示形式由m个连续的1组成的最大数，然后由m-1个连续的零组成，并且没有其他值
例子：
Input : n = 7
Output : 6
Explanation: 6's binary representation is 110,
and 7's is 111, so 6 consists of 2 consecutive
1's and then 1 consecutive 0.
Input : 130
Output : 120
Explanation: 28 and 120 are the only numbers <=120,
28 is 11100 consists of 3 consecutive 1's and then
2 consecutive 0's. 120 is 1111000 consists of 4
consecutive 1's and then 3 consecutive 0's. So 120
is the greatest of number<=120 which meets the
given condition.
天真的方法是从1遍历到N，并检查由m个连续的1和m-1个连续的0组成的每个二进制表示形式，并存储满足给定条件的最大二进制表示形式。
一种有效的方法是观察数字模式，
[ 1(1)，6(110)，28(11100)，120(1111000)，496(111110000)，…。]
为了得到满足条件的数字的公式，我们以120为例-
120表示为1111000，其中m = 4 1和m = 3 0。将1111000转换为十进制，我们得到：
2 ^ 3 + 2 ^ 4 + 2 ^ 5 + 2 ^ 6，可以表示为(2 ^ m-1 + 2 ^ m + 2 ^ m + 1 +…2 ^ m + 2，2 ^ 2 * m)
2 ^ 3 *(1 + 2 + 2 ^ 2 + 2 ^ 3)可以表示为(2 ^(m-1)*(1 + 2 + 2 ^ 2 + 2 ^ 3 + .. 2 ^(m -1))
2 ^ 3 *(2 ^ 4-1)可以表示为[2 ^(m-1)*(2 ^ m -1)]。
因此，满足给定条件的所有数字都可以表示为
[2^(m-1) * (2^m -1)]
我们可以迭代直到数字不超过N并打印所有可能的元素中的最大元素。仔细观察会发现，在m = 33时，它将超过10 ^ 18标记，因此我们正在计算单位时间中的数字，因为log(32)接近计算pow所需的常数。
因此，总体复杂度将为O(1)。
C++
// CPP program to find largest number
// smaller than equal to n with m set
// bits then m-1 0 bits.
#include
using namespace std;
// Returns largest number with m set
// bits then m-1 0 bits.
long long answer(long long n)
{
// Start with 2 bits.
long m = 2;
// initial answer is 1
// which meets the given condition
long long ans = 1;
long long r = 1;
// check for all numbers
while (r < n) {
// compute the number
r = (int)(pow(2, m) - 1) * (pow(2, m - 1));
// if less then N
if (r < n)
ans = r;
// increment m to get the next number
m++;
}
return ans;
}
// driver code to check the above condition
int main()
{
long long n = 7;
cout << answer(n);
return 0;
} Java
// java program to find largest number
// smaller than equal to n with m set
// bits then m-1 0 bits.
public class GFG {
// Returns largest number with
// m set bits then m-1 0 bits.
static long answer(long n)
{
// Start with 2 bits.
long m = 2;
// initial answer is 1 which
// meets the given condition
long ans = 1;
long r = 1;
// check for all numbers
while (r < n) {
// compute the number
r = ((long)Math.pow(2, m) - 1) *
((long)Math.pow(2, m - 1));
// if less then N
if (r < n)
ans = r;
// increment m to get
// the next number
m++;
}
return ans;
}
// Driver code
public static void main(String args[]) {
long n = 7;
System.out.println(answer(n));
}
}
// This code is contributed by Sam007Python3
# Python3 program to find
# largest number smaller
# than equal to n with m
# set bits then m-1 0 bits.
import math
# Returns largest number
# with m set bits then
# m-1 0 bits.
def answer(n):
# Start with 2 bits.
m = 2;
# initial answer is
# 1 which meets the
# given condition
ans = 1;
r = 1;
# check for all numbers
while r < n:
# compute the number
r = (int)((pow(2, m) - 1) *
(pow(2, m - 1)));
# if less then N
if r < n:
ans = r;
# increment m to get
# the next number
m = m + 1;
return ans;
# Driver Code
print(answer(7));
# This code is contributed by mits.C#
// C# program to find largest number
// smaller than equal to n with m set
// bits then m-1 0 bits.
using System;
class GFG {
// Returns largest number with
// m set bits then m-1 0 bits.
static long answer(long n)
{
// Start with 2 bits.
long m = 2;
// initial answer is 1 which
// meets the given condition
long ans = 1;
long r = 1;
// check for all numbers
while (r < n) {
// compute the number
r = ((long)Math.Pow(2, m) - 1) *
((long)Math.Pow(2, m - 1));
// if less then N
if (r < n)
ans = r;
// increment m to get
// the next number
m++;
}
return ans;
}
// Driver Code
static public void Main ()
{
long n = 7;
Console.WriteLine(answer(n));
}
}
// This code is contributed by vt_m.PHP
输出：
6