最小化系列 a, a/b^1, a/b^2, a/b^3, ..., a/b^n 中 a 的值，使得初始非零项的总和至少变为 S

给定两个整数b和S 。任务是找到“ a ”的最小值，使得对于初始非零项的总和等于或大于“ S ”。

a, a/b¹, a/b², a/b³, …………., a/bⁿ

例子：

Input: b = 2, S = 4
Output: 3
Explanation: 

• Let a = 1, S = 1/2⁰ + 1/2¹ = 1 + 0 = 1 < 4.
• Let a =2, S = 2/2⁰ + 2/2¹ + 2/2² = 2 + 1 + 0 = 3 < 4.
• Let a = 3, S = 3/2⁰ + 3/2¹ + 3/2² = 3 + 1 + 0 = 4 = S.

So, a = 3 is the answer.

Input: b = 8, S = 25
Output: 23

方法：这个问题可以通过二分查找来解决。显然，如果数字“ a ”是一个答案，那么每个数字n > a也是一个答案，因为值只会变得更多，但我们需要找到最小值。因此，要检查某个数字“a”，我们可以使用问题本身中给出的公式。请按照以下步骤解决问题：

• 将变量a初始化为1，低为0 ，高为S。
• 在 while 循环中遍历直到low小于等于high并执行以下任务：
  • 将变量mid初始化为低和高的平均值。
  • 将x初始化为b并将sum初始化为mid。
  • 在while循环中遍历直到mid/x大于0并执行以下任务：
    • 将mid/x的值添加到变量sum中。
    • 将值b乘以变量x。
  • 如果sum大于等于S ，则将a设置为mid ，将high设置为mid-1。
  • 否则设置低为中+1。
• 执行上述步骤后，打印a的值作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum value
// of numerator such that sum of certain
// terms in the given series become
// equal or greater than X
int findMinNumerator(int b, int S)
{
 
    // Variable to store the ans
    // initialized with 1 which
    // can be the minimum answer
    int a = 1;
    int low = 0, high = S;
 
    // Iterate till low is less or
    // equal to high
    while (low <= high) {
 
        // Find the mid value
        int mid = (low + high) / 2;
        int x = b, sum = mid;
 
        // While mid / x is greater than
        // 0 keep updating sum and x
        while (mid / x > 0) {
            sum += mid / x;
            x *= b;
        }
 
        // If sum is greater than S,
        // store mid in ans and update
        // high to search other minimum
        if (sum >= S) {
            a = mid;
            high = mid - 1;
        }
 
        // Else update low as (mid + 1)
        else if (sum < S) {
            low = mid + 1;
        }
    }
 
    // Return the answer
    return a;
}
 
// Driver Code
int main()
{
    int b = 2, S = 4;
 
    cout << findMinNumerator(b, S);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
public class GFG
{
// Function to find the minimum value
// of numerator such that sum of certain
// terms in the given series become
// equal or greater than X
static int findMinNumerator(int b, int S)
{
     
    // Variable to store the ans
    // initialized with 1 which
    // can be the minimum answer
    int a = 1;
    int low = 0, high = S;
 
    // Iterate till low is less or
    // equal to high
    while (low <= high) {
 
        // Find the mid value
        int mid = (low + high) / 2;
        int x = b, sum = mid;
 
        // While mid / x is greater than
        // 0 keep updating sum and x
        while (mid / x > 0) {
            sum += mid / x;
            x *= b;
        }
 
        // If sum is greater than S,
        // store mid in ans and update
        // high to search other minimum
        if (sum >= S) {
            a = mid;
            high = mid - 1;
        }
 
        // Else update low as (mid + 1)
        else if (sum < S) {
            low = mid + 1;
        }
    }
 
    // Return the answer
    return a;
}
 
// Driver Code
public static void main(String args[])
{
    int b = 2, S = 4;
    System.out.println(findMinNumerator(b, S));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python 3 program for the above approach
 
# Function to find the minimum value
# of numerator such that sum of certain
# terms in the given series become
# equal or greater than X
def findMinNumerator(b, S):
 
    # Variable to store the ans
    # initialized with 1 which
    # can be the minimum answer
    a = 1
    low = 0
    high = S
 
    # Iterate till low is less or
    # equal to high
    while (low <= high):
 
        # Find the mid value
        mid = (low + high) // 2
        x = b
        sum = mid
 
        # While mid / x is greater than
        # 0 keep updating sum and x
        while (mid // x > 0):
            sum += mid // x
            x *= b
 
        # If sum is greater than S,
        # store mid in ans and update
        # high to search other minimum
        if (sum >= S):
            a = mid
            high = mid - 1
 
        # Else update low as (mid + 1)
        elif (sum < S):
            low = mid + 1
 
    # Return the answer
    return a
 
# Driver Code
if __name__ == "__main__":
 
    b = 2
    S = 4
 
    print(findMinNumerator(b, S))
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
using System.Collections;
 
public class GFG
{
// Function to find the minimum value
// of numerator such that sum of certain
// terms in the given series become
// equal or greater than X
static int findMinNumerator(int b, int S)
{
     
    // Variable to store the ans
    // initialized with 1 which
    // can be the minimum answer
    int a = 1;
    int low = 0, high = S;
 
    // Iterate till low is less or
    // equal to high
    while (low <= high) {
 
        // Find the mid value
        int mid = (low + high) / 2;
        int x = b, sum = mid;
 
        // While mid / x is greater than
        // 0 keep updating sum and x
        while (mid / x > 0) {
            sum += mid / x;
            x *= b;
        }
 
        // If sum is greater than S,
        // store mid in ans and update
        // high to search other minimum
        if (sum >= S) {
            a = mid;
            high = mid - 1;
        }
 
        // Else update low as (mid + 1)
        else if (sum < S) {
            low = mid + 1;
        }
    }
 
    // Return the answer
    return a;
}
 
// Driver Code
public static void Main()
{
    int b = 2, S = 4;
    Console.Write(findMinNumerator(b, S));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

 
        // JavaScript Program to implement
        // the above approach
 
        // Function to find the minimum value
        // of numerator such that sum of certain
        // terms in the given series become
        // equal or greater than X
        function findMinNumerator(b, S) {
 
            // Variable to store the ans
            // initialized with 1 which
            // can be the minimum answer
            let a = 1;
            let low = 0, high = S;
 
            // Iterate till low is less or
            // equal to high
            while (low <= high) {
 
                // Find the mid value
                let mid = Math.floor((low + high) / 2);
                let x = b, sum = mid;
 
                // While mid / x is greater than
                // 0 keep updating sum and x
                while (Math.floor(mid / x) > 0) {
                    sum += mid / x;
                    x *= b;
                }
 
                // If sum is greater than S,
                // store mid in ans and update
                // high to search other minimum
                if (sum >= S) {
                    a = mid;
                    high = mid - 1;
                }
 
                // Else update low as (mid + 1)
                else if (sum < S) {
                    low = mid + 1;
                }
            }
 
            // Return the answer
            return a;
        }
 
        // Driver Code
        let b = 2, S = 4;
        document.write(findMinNumerator(b, S));
 
    // This code is contributed by Potta Lokesh
    

输出

3

时间复杂度： O(log ² N)
辅助空间： O(1)