找到可被 a 或 b 或 c 整除的第 n 项
原文:https://www . geeksforgeeks . org/find-第 n 个术语-可被-a 或-b 或-c 除尽/
给定四个整数 a 、 b 、 c 和 N 。任务是找到可被 a 、 b 或 c 整除的 N 第 项。 例:
输入: a = 2，b = 3，c = 5，N = 10 输出: 14 序列为 2，3，4，5，6，8，9，10，12，14，15，16…… 输入: a = 3，b = 5，c = 7，N = 10 输出: 18
天真方法:一个简单的方法是遍历所有术语，从 1 开始，直到我们找到所需的 N th 术语，该术语可被Ab或 c 整除。该解决方案的时间复杂度为 0(N)。 高效途径:思路是用二分搜索法。在这里，我们可以通过使用以下公式来计算从 1 到数有多少个数可以被 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)
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
// Rajnis09
计算机编程语言
# 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 29
C
// 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 */
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return
// gcd of a and b
function gcd(a, b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to return the lcm of a and b
function lcm(a, b)
{
return parseInt((a * b) / gcd(a, b));
}
// Function to return the count of numbers
// from 1 to num which are divisible by a, b or c
function 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 (parseInt(num / a) + parseInt(num / b) +
parseInt(num / c)
- parseInt(num / lcm(a, b))
- parseInt(num / lcm(b, c))
- parseInt(num / lcm(a, c))
+ parseInt(num / lcm(a, lcm(b, c))));
}
// Function to find the nth term
// divisible by a, b or c
// by using binary search
function findNthTerm(a, b, c, n)
{
// Set low to 1 and high to max (a, b, c) * n
let low = 1, high = Number.MAX_VALUE, mid;
while (low < high) {
mid = low + parseInt((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
let a = 2, b = 3, c = 5, n = 10;
document.write(findNthTerm(a, b, c, n));
</script>
Output:
14
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