给定四个整数a ， b ， c和N。任务是找到第n个可被a ， b或c整除的项。
例子：
Input: a = 2, b = 3, c = 5, N = 10
Output: 14
Sequence is 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16…
Input: a = 3, b = 5, c = 7, N = 10
Output: 18
天真的方法：一种简单的方法是遍历从1开始的所有项，直到找到所需的第N个项，该项可以被a ， b或c整除。该解决方案的时间复杂度为O(N)。
高效的方法：这个想法是使用二进制搜索。在这里，我们可以使用以下公式计算从1到num的数可以被a ， b或c整除： (num / a)+(num / b)+(num / c)–(num / lcm(a， b))–(num / lcm(b，c))–(num / lcm(a，c))+(num / lcm(a，b，c)))
上面的公式是使用集合理论推导的
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return
// gcd of a and b
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to return the lcm of a and b
int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
// Function to return the count of numbers
// from 1 to num which are divisible by a, b or c
int divTermCount(int a, int b, int c, int num)
{
// Calculate number of terms divisible by a and
// by b and by c then, remove the terms which is are
// divisible by both a and b, both b and c, both
// c and a and then add which are divisible by a and
// b and c
return ((num / a) + (num / b) + (num / c)
- (num / lcm(a, b))
- (num / lcm(b, c))
- (num / lcm(a, c))
+ (num / lcm(a, lcm(b, c))));
}
// Function to find the nth term
// divisible by a, b or c
// by using binary search
int findNthTerm(int a, int b, int c, int n)
{
// Set low to 1 and high to max (a, b, c) * n
int low = 1, high = INT_MAX, mid;
while (low < high) {
mid = low + (high - low) / 2;
// If the current term is less than
// n then we need to increase low
// to mid + 1
if (divTermCount(a, b, c, mid) < n)
low = mid + 1;
// If current term is greater than equal to
// n then high = mid
else
high = mid;
}
return low;
}
// Driver code
int main()
{
int a = 2, b = 3, c = 5, n = 10;
cout << findNthTerm(a, b, c, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to return
// gcd of a and b
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to return the lcm of a and b
static int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
// Function to return the count of numbers
// from 1 to num which are divisible by a, b or c
static int divTermCount(int a, int b, int c, int num)
{
// Calculate number of terms divisible by a and
// by b and by c then, remove the terms which is are
// divisible by both a and b, both b and c, both
// c and a and then add which are divisible by a and
// b and c
return ((num / a) + (num / b) + (num / c)
- (num / lcm(a, b))
- (num / lcm(b, c))
- (num / lcm(a, c))
+ (num / lcm(a, lcm(b, c))));
}
// Function to find the nth term
// divisible by a, b or c
// by using binary search
static int findNthTerm(int a, int b, int c, int n)
{
// Set low to 1 and high to max (a, b, c) * n
int low = 1, high = Integer.MAX_VALUE, mid;
while (low < high) {
mid = low + (high - low) / 2;
// If the current term is less than
// n then we need to increase low
// to mid + 1
if (divTermCount(a, b, c, mid) < n)
low = mid + 1;
// If current term is greater than equal to
// n then high = mid
else
high = mid;
}
return low;
}
// Driver code
public static void main(String[] args)
{
int a = 2, b = 3, c = 5, n = 10;
System.out.println(findNthTerm(a, b, c, n));
}
}
// This code is contributed by
// Rajnis09Python
# Python3 implementation of the approach
# Function to return
# gcd of a and b
def gcd(a, b):
if (a == 0):
return b
return gcd(b % a, a)
# Function to return the lcm of a and b
def lcm(a, b):
return ((a * b) // gcd(a, b))
# Function to return the count of numbers
# from 1 to num which are divisible by a, b or c
def divTermCount(a, b, c, num):
# Calculate number of terms divisible by a and
# by b and by c then, remove the terms which is are
# divisible by both a and b, both b and c, both
# c and a and then add which are divisible by a and
# b and c
return ((num // a) + (num // b) + (num // c)
- (num // lcm(a, b))
- (num // lcm(b, c))
- (num // lcm(a, c))
+ (num // lcm(a, lcm(b, c))))
# Function to find the nth term
# divisible by a, b or c
# by using binary search
def findNthTerm(a, b, c, n):
# Set low to 1 and high to max (a, b, c) * n
low = 1
high = 10**9
mid=0
while (low < high):
mid = low + (high - low) // 2
# If the current term is less than
# n then we need to increase low
# to mid + 1
if (divTermCount(a, b, c, mid) < n):
low = mid + 1
# If current term is greater than equal to
# n then high = mid
else:
high = mid
return low
# Driver code
a = 2
b = 3
c = 5
n = 10
print(findNthTerm(a, b, c, n))
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return
// gcd of a and b
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to return the lcm of a and b
static int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
// Function to return the count of numbers
// from 1 to num which are divisible by a, b or c
static int divTermCount(int a, int b, int c, int num)
{
// Calculate number of terms divisible by a and
// by b and by c then, remove the terms which is are
// divisible by both a and b, both b and c, both
// c and a and then add which are divisible by a and
// b and c
return ((num / a) + (num / b) + (num / c)
- (num / lcm(a, b))
- (num / lcm(b, c))
- (num / lcm(a, c))
+ (num / lcm(a, lcm(b, c))));
}
// Function to find the nth term
// divisible by a, b or c
// by using binary search
static int findNthTerm(int a, int b, int c, int n)
{
// Set low to 1 and high to max (a, b, c) * n
int low = 1, high = int.MaxValue, mid;
while (low < high) {
mid = low + (high - low) / 2;
// If the current term is less than
// n then we need to increase low
// to mid + 1
if (divTermCount(a, b, c, mid) < n)
low = mid + 1;
// If current term is greater than equal to
// n then high = mid
else
high = mid;
}
return low;
}
// Driver code
public static void Main(String[] args)
{
int a = 2, b = 3, c = 5, n = 10;
Console.WriteLine(findNthTerm(a, b, c, n));
}
}
/* This code is contributed by PrinciRaj1992 */输出：
14