📜  给定条件下几何级数中的项数

📅  最后修改于: 2021-06-26 11:15:49             🧑  作者: Mango

几何级数是整数b1,b2,b3,…的序列,其中对于每个i> 1,相应的项均满足条件b i = b i-1 * q,其中q被称为级数的公共比率。

给定由两个整数b1和q定义的几何级数b,以及m个“坏”整数a1,a2,..,am和整数l定义的所有条件项| b i |满足<= l(| x |表示x的绝对值)。计算我们的序列中将有多少个数字,或者在无限多个整数的情况下显示“ inf”。
注意:如果一项等于“不良”整数之一,请跳过该术语并前进至下一项。

例子:

Input : b1 = 3, q = 2, l = 30,
        m = 4 
        6 14 25 48
Output : 3
The progression will be 3 12 24. 
6 will also be there but because
it is a bad integer we won't include it

Input : b1 = 123, q = 1, l = 2143435
        m = 4
        123 11 -5453 141245
Output : 0
As value of q is 1, progression will 
always be 123 and would become infinity
but because it is a bad integer we
won't include it and hence our value
will become 0

Input : b1 = 123, q = 1, l = 2143435 
        m = 4
        5234 11 -5453 141245
Output : inf
In this case, value will be infinity 
because series will always be 123 as 
q is 1 and 123 is not a bad integer.

方法:
我们可以在不同情况下划分解决方案:
情况1:如果系列的起始值大于给定的限制,则输出为0。
情况2:如果系列或q的起始值为0,则还有另外三种情况:
情况2.a:如果未将0作为错误整数给出,则答案将变为inf。
情况2.b:如果b1!= 0但q为0且b1也不是错误的整数,则答案将变为1。
情况2.c:如果将0作为错误整数给出且b1 = 0,则答案将变为0。
情况3:如果q = 1,我们将检查b1是否为错误整数。如果是,则答案将为0,否则答案将为inf。
情况4:如果q = -1,请检查b1和-b1是否存在,如果存在,我们的答案将为0,否则我们的答案将为inf。
情况5:如果以上情况均不成立,则只需对b1循环运行直到l并计算元素数。

下面是上述方法的实现:

C++
// CPP program to find number of terms  
// in Geometric Series
#include 
using namespace std;
  
// A map to keep track of the bad integers
map mapp; 
  
// Function to calculate No. of elements
// in our series
void progression(int b1, int q, int l,
                 int m, int bad[])
{
    // Updating value of our map
    for (int i = 0; i < m; i++) 
        mapp[bad[i]] = 1;    
      
    // if starting value is greate
    // r than our given limit
    if (abs(b1) > l) 
        cout << "0";
          
    // if q or starting value is 0    
    else if (q == 0 || b1 == 0) 
    {   
        // if 0 is not a bad integer,
        // answer becomes inf
        if (mapp[0] != 1)         
            cout << "inf";
          
        // if q is 0 and b1 is not and b1
        // is not a bad integer, answer becomes 1
        else if (mapp[0] == 1 && mapp[b1] != 1) 
            cout << "1";
  
        else // else if 0 is bad integer and
            // b1 is also a bad integer,
            // answer becomes 0
            cout << "0";
    }
    else if (q == 1) // if q is 1
    {   
        // and b1 is not a bad integer,
        // answer becomes inf
        if (mapp[b1] != 1) 
            cout << "inf";
  
        else // else answer is 0
            cout << "0";
    }
    else if (q == -1) // if q is -1
    {   
        // and either b1 or -b1 is not 
        // present answer becomes inf
        if (mapp[b1] != 1 || mapp[-1 * b1] != 1) 
            cout << "inf";
  
        else // else answer becomes 0
            cout << "0";
    }
    else // if none of the above case is true, 
         // simpy calculate the number of 
        // elements in our series
    {
        int co = 0;
        while (abs(b1) <= l) {
            if (mapp[b1] != 1)
                co++;
            b1 *= 1LL * q;
        }
        cout << co;
    }
}
  
// driver code
int main()
{   
    // starting value of series,
    // number to be multiplied,
    // limit within which our series,
    // No. of bad integers given
    int b1 = 3, q = 2, l = 30, m = 4;
      
    // Bad integers
    int bad[4] = { 6, 14, 25, 48 }; 
      
    progression(b1, q, l, m, bad);
      
    return 0;
}


Java
// Java program to find number of terms 
// in Geometric Series
import java.util.*;
  
class GFG 
{
  
    // A map to keep track of the bad integers
    static HashMap map = new HashMap<>();
  
    // Function to calculate No. of elements
    // in our series
    static void progression(int b1, int q, int l,
                                int m, int[] bad) 
    {
  
        // Updating value of our map
        for (int i = 0; i < m; i++)
            map.put(bad[i], true);
  
        // if starting value is greate
        // r than our given limit
        if (Math.abs(b1) > l)
            System.out.print("0");
  
        // if q or starting value is 0
        else if (q == 0 || b1 == 0)
        {
  
            // if 0 is not a bad integer,
            // answer becomes inf
            if (!map.containsKey(0))
                System.out.print("inf");
  
            // if q is 0 and b1 is not and b1
            // is not a bad integer, answer becomes 1
            else if (map.get(0) == true && !map.containsKey(b1))
                System.out.print("1");
  
            // else if 0 is bad integer and
            // b1 is also a bad integer,
            // answer becomes 0
            else
                System.out.print("0");
        }
  
        // if q is 1
        else if (q == 1) 
        {
  
            // and b1 is not a bad integer,
            // answer becomes inf
            if (!map.containsKey(b1))
                System.out.print("inf");
  
            // else answer is 0
            else
                System.out.print("0");
  
        }
  
        // if q is -1
        else if (q == -1)
        {
  
            // and either b1 or -b1 is not
            // present answer becomes inf
            if (!map.containsKey(b1) || !map.containsKey(-1 * b1))
                System.out.print("inf");
  
            // else answer becomes 0
            else
                System.out.print("0");
  
              
        } 
          
        // if none of the above case is true,
        // simpy calculate the number of
        // elements in our series
        else 
        {
            int co = 0;
            while (Math.abs(b1) <= l) 
            {
                if (!map.containsKey(b1))
                    co++;
                b1 *= q;
            }
            System.out.print(co);
        }
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        // starting value of series,
        // number to be multiplied,
        // limit within which our series,
        // No. of bad integers given
        int b1 = 3, q = 2, l = 30, m = 4;
  
        // Bad integers
        int[] bad = { 6, 14, 25, 48 };
  
        progression(b1, q, l, m, bad);
    }
}
  
// This code is contributed by
// sanjeev2552


Python3
# Python3 program to find number of terms
# in Geometric Series
  
# A map to keep track of the bad integers
mpp=dict()
  
# Function to calculate No. of elements
# in our series
def progression(b1, q, l, m, bad):
  
    # Updating value of our map
    for i in range(m):
        mpp[bad[i]] = 1
  
    # if starting value is greate
    # r than our given limit
    if (abs(b1) > l):
        print("0",end="")
          
    # if q or starting value is 0
    elif (q == 0 or b1 == 0) :
          
        # if 0 is not a bad integer,
        # answer becomes inf
        if (0 not in mpp.keys()):
            print("inf",end="")
              
        # if q is 0 and b1 is not and b1
        # is not a bad integer, answer becomes 1
        elif (mpp[0] == 1 and b1 not in mpp.keys()) :
            print("1",end="")
  
        else:
            # b1 is also a bad integer,
            # answer becomes 0
            print("0",end="")
    elif (q == 1): # if q is 1
        # and b1 is not a bad integer,
        # answer becomes inf
        if (b1 not in mpp.keys()) :
            print("inf",end="")
        else: # else answer is 0
            print("0",end="")
  
    elif (q == -1): # if q is -1
        # and either b1 or -b1 is not
        # present answer becomes inf
        if (b1 not in mpp.keys() or -1 * b1 not in mpp.keys()) :
            print("inf",end="")
  
        else :# else answer becomes 0
            print("0",end="")
    else :# if none of the above case is true,
        # simpy calculate the number of
        # elements in our series
        co = 0
        while (abs(b1) <= l):
            if (b1 not in mpp.keys()):
                co+=1
            b1 *= q
        print(co,end="")
  
  
# Driver code
  
# starting value of series,
# number to be multiplied,
# limit within which our series,
# No. of bad integers given
b1 = 3
q = 2
l = 30
m = 4
  
# Bad integers
bad=[6, 14, 25, 48]
  
progression(b1, q, l, m, bad)
  
# This code is contributed by mohit kumar 29


C#
// C# program to find number of terms 
// in Geometric Series
using System;
using System.Collections.Generic;
  
class GFG 
{
  
    // A map to keep track of the bad integers
    static Dictionary map = new Dictionary();
  
    // Function to calculate No. of elements
    // in our series
    static void progression(int b1, int q, int l,
                            int m, int[] bad) 
    {
  
        // Updating value of our map
        for (int i = 0; i < m; i++)
            if(!map.ContainsKey(bad[i]))
                map.Add(bad[i], true);
  
        // if starting value is greate
        // r than our given limit
        if (Math.Abs(b1) > l)
            Console.Write("0");
  
        // if q or starting value is 0
        else if (q == 0 || b1 == 0)
        {
  
            // if 0 is not a bad integer,
            // answer becomes inf
            if (!map.ContainsKey(0))
                Console.Write("inf");
  
            // if q is 0 and b1 is not and b1
            // is not a bad integer, answer becomes 1
            else if (map[0] == true && 
                    !map.ContainsKey(b1))
                Console.Write("1");
  
            // else if 0 is bad integer and
            // b1 is also a bad integer,
            // answer becomes 0
            else
                Console.Write("0");
        }
  
        // if q is 1
        else if (q == 1) 
        {
  
            // and b1 is not a bad integer,
            // answer becomes inf
            if (!map.ContainsKey(b1))
                Console.Write("inf");
  
            // else answer is 0
            else
                Console.Write("0");
        }
  
        // if q is -1
        else if (q == -1)
        {
  
            // and either b1 or -b1 is not
            // present answer becomes inf
            if (!map.ContainsKey(b1) || 
                !map.ContainsKey(-1 * b1))
                Console.Write("inf");
  
            // else answer becomes 0
            else
                Console.Write("0");
        } 
          
        // if none of the above case is true,
        // simpy calculate the number of
        // elements in our series
        else
        {
            int co = 0;
            while (Math.Abs(b1) <= l) 
            {
                if (!map.ContainsKey(b1))
                    co++;
                b1 *= q;
            }
            Console.Write(co);
        }
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        // starting value of series,
        // number to be multiplied,
        // limit within which our series,
        // No. of bad integers given
        int b1 = 3, q = 2, l = 30, m = 4;
  
        // Bad integers
        int[] bad = { 6, 14, 25, 48 };
  
        progression(b1, q, l, m, bad);
    }
}
  
// This code is contributed by Rajput-Ji


输出:

3