在给定的单调序列中查找元素位置

给定一个整数k和一个单调递增的序列：
f(n) = an + bn [log2(n)] + cn^3其中 ( a = 1, 2, 3, …), ( b = 1, 2, 3, …), ( c = 0, 1, 2、3、……)
这里，[log ₂ (n)] 的意思是，将 log 取到基数 2 并将值向下舍入 因此，
如果 n = 1，则值为 0。
如果 n = 2-3，则值为 1。
如果 n = 4-7，则值为 2。
如果 n = 8-15，则值为 3。
任务是找到值n使得f(n) = k ，如果k不属于序列，则打印0 。

注：数值以64位表示，a、b、c三个整数不超过100。

例子：

Input: a = 2, b = 1, c = 1, k = 12168587437017 
Output: 23001 
f(23001) = 12168587437017

Input: a = 7, b = 3, c = 0, k = 119753085330 
Output: 1234567890 

朴素方法：给定 a、b、c 的值，为 n 的每个值找到 f(n) 的值并进行比较。

高效方法：使用二分搜索，选择n = (min + max) / 2其中min和max是n可能的最小值和最大值，然后，

• 如果f(n) < k则增加n 。
• 如果f(n) > k则递减n 。
• 如果f(n) = k则n是必需的答案。
• 重复上述步骤，直到找到所需值或在序列中不可能。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
#include 
#define SMALL_N 1000000
#define LARGE_N  1000000000000000
using namespace std;
 
// Function to return the value of f(n) for given values of a, b, c, n
long long func(long long a, long long b, long long c, long long n)
{
    long long res = a * n;
    long long logVlaue = floor(log2(n));
    res += b * n * logVlaue;
    res += c * (n * n * n);
    return res;
}
 
long long getPositionInSeries(long long a, long long b,
                             long long c, long long k)
{
    long long start = 1, end = SMALL_N;
 
    // if c is 0, then value of n can be in order of 10^15.
    // if c!=0, then n^3 value has to be in order of 10^18
    // so maximum value of n can be 10^6.
    if (c == 0) {
        end = LARGE_N;
    }
    long long ans = 0;
 
    // for efficient searching, use binary search.
    while (start <= end) {
        long long mid = (start + end) / 2;
        long long val = func(a, b, c, mid);
        if (val == k) {
            ans = mid;
            break;
        }
        else if (val > k) {
            end = mid - 1;
        }
        else {
            start = mid + 1;
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    long long a = 2, b = 1, c = 1;
    long long k = 12168587437017;
 
    cout << getPositionInSeries(a, b, c, k);
 
    return 0;
}

Python3

# Python 3 implementation of the approach
from math import log2, floor
SMALL_N = 1000000
LARGE_N = 1000000000000000
 
# Function to return the value of f(n)
# for given values of a, b, c, n
def func(a, b, c, n) :
     
    res = a * n
    logVlaue = floor(log2(n))
    res += b * n * logVlaue
    res += c * (n * n * n)
    return res
 
def getPositionInSeries(a, b, c, k) :
     
    start = 1
    end = SMALL_N
 
    # if c is 0, then value of n
    # can be in order of 10^15.
    # if c!=0, then n^3 value has
    # to be in order of 10^18
    # so maximum value of n can be 10^6.
    if (c == 0) :
        end = LARGE_N
     
    ans = 0
 
    # for efficient searching,
    # use binary search.
    while (start <= end) :
         
        mid = (start + end) // 2
        val = func(a, b, c, mid)
        if (val == k) :
            ans = mid
            break
     
        elif (val > k) :
            end = mid - 1
 
        else :
            start = mid + 1
         
    return ans;
 
# Driver code
if __name__ == "__main__" :
     
    a = 2
    b = 1
    c = 1
    k = 12168587437017
 
    print(getPositionInSeries(a, b, c, k))
 
# This code is contributed by Ryuga

PHP

 $k)
            $end = $mid - 1;
 
        else
            $start = $mid + 1;
    }
    return $ans;
}
 
// Driver code
$a = 2;
$b = 1;
$c = 1;
$k = 12168587437017;
 
print(getPositionInSeries($a, $b, $c, $k));
 
// This code is contributed by mits
?>

Javascript

 
// Javascript implementation of the approach
const SMALL_N = 1000000;
const LARGE_N = 1000000000000000;
 
// Function to return the value of f(n)
// for given values of a, b, c, n
function func(a, b, c, n)
{
    let res = a * n;
    let logVlaue = Math.floor(Math.log(n) /
                              Math.log(2));
    res += b * n * logVlaue;
    res += c * (n * n * n);
    return res;
}
 
function getPositionInSeries(a, b, c, k)
{
    let start = 1, end = SMALL_N;
 
    // If c is 0, then value of n can be
    // in order of 10^15. If c!=0, then
    // n^3 value has to be in order of 10^18
    // so maximum value of n can be 10^6.
    if (c == 0)
    {
        end = LARGE_N;
    }
    let ans = 0;
 
    // For efficient searching, use binary search.
    while (start <= end)
    {
        let mid = parseInt((start + end) / 2);
        let val = func(a, b, c, mid);
        if (val == k)
        {
            ans = mid;
            break;
        }
        else if (val > k)
        {
            end = mid - 1;
        }
        else
        {
            start = mid + 1;
        }
    }
    return ans;
}
 
// Driver code
let a = 2, b = 1, c = 1;
let k = 12168587437017;
 
document.write(getPositionInSeries(a, b, c, k));
 
// This code is contributed by souravmahato348
 

输出：

23001

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