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 29C#
// 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