给定整数N 以及两个整数D1和D2 ( <10 )，任务是找到小于或等于N的仅由数字D1和D2组成的互质对数。
例子：
Input: N = 30, D1 = 2, D2 = 3
Output: 5
Explanation:
All possible pairs of numbers upto 30 having digits 2 and 3 which are co-prime to each other are (2, 3), (2, 23), (3, 22), (3, 23), (22, 23) .
Input: N = 100, D1 = 5, D2 = 6
Output: 8
Explanation:
All possible pairs of numbers upto 100 having digits 5 and 6 which are co-prime to each other are (5, 6), (5, 56), (5, 66), (6, 55), (6, 65), (55, 56), (56, 65), (65, 66).
方法：解决此问题的想法是基于以下观察结果：
Observation:
- Find and append every number consisting only given two digits which are less than or equal to N into a list.
- Now it’s easy to find the number of unordered pairs which are co-primes.
- Note that there can be maximum 1 + 2 + 22 + 23 + 24 + ………210= 2047 numbers in the list i.e. Overall Time Complexity cannot exceed O(2047 * 2047), as the maximum no. of digits possible is 9.
请按照以下步骤解决问题：
- 初始化一个空列表，说l ， 并将给定的两位数字附加为 列表中的一个字符串。
- 排序列表。
- 初始化两个列表，例如total和temp2，以供进一步使用。
- 迭代直到l [0] <10：
- 将给定的两位数字作为字符串附加到列表l中存在的所有元素。
- 按照以下所示的方式继续更新l ：
- [2，3]-> [‘2’+’2’，’2’+’3’，’3’+’2’，’3’+’3’]
- [22，23，32，33] –> [’22’+’2’，’22’+’3’，’23’+’2’，’23’+’3’，’32’+’2 ‘，’32’+’3’，’33’+’2’，’33’+’3’]等。
- 定义一个函数numOfPairs() ，该函数返回无序互质对数的计数。
- 打印以上函数返回的计数作为答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check whether given
// integers are co-prime or not
int coprime(int a, int b) { return (__gcd(a, b) == 1); }
// Utility function to count
// number of co-prime pairs
int numOfPairs(vector arr, int N)
{
int count = 0;
// Traverse the array
for (int i = 0; i < N - 1; i++) {
for (int j = i + 1; j < N; j++) {
// If co-prime
if (coprime(stoi(arr[i]), stoi(arr[j]))) {
// Increment count
count = count + 1;
}
}
}
// Return count
return count;
}
// Function to count number
// of co-prime pairs
int noOfCoPrimePairs(int N, int d1, int d2)
{
// Stores digits in string form
vector l;
l.push_back(to_string(d1));
l.push_back(to_string(d2));
// Sort the list
sort(l.begin(), l.end());
if (N < stoi(l[1]))
return 0;
// Keep two copies of list l
vector total = l;
vector temp2 = l;
int flag = 0;
vector temp3;
// Generate 2 digit numbers
// using d1 and d2
while (l[0].length() < 10) {
for (int i = 0; i < l.size(); i++) {
for (int j = 0; j < 2; j++) {
// If current number
// does not exceed N
if (stoi(l[i] + temp2[j]) > N) {
flag = 1;
break;
}
total.push_back(l[i] + temp2[j]);
temp3.push_back(l[i] + temp2[j]);
}
if (flag == 1)
break;
}
if (flag == 1)
break;
l = temp3;
vector temp3;
}
// Stores length of list
int lenOfTotal = total.size();
// Stores number of co-prime pairs
int ans = numOfPairs(total, lenOfTotal);
// Print number of co-prime pairs
cout << (ans);
}
// Driver Code
int main()
{
// Given value of N, d1, d2
int N = 30, d1 = 2, d2 = 3;
// Function call to count
// number of co-prime pairs
noOfCoPrimePairs(N, d1, d2);
}
// This code is contributed by ukasp. Java
// Java program for the above approach
import java.util.*;
class GFG{
static int GCD(int a, int b)
{
return b == 0 ? a : GCD(b, a % b);
}
// Function to check whether given
// integers are co-prime or not
static boolean coprime(int a, int b)
{
if (GCD(a, b) == 1)
return true;
return false;
}
// Utility function to count
// number of co-prime pairs
static int numOfPairs(ArrayList arr, int N)
{
int count = 0;
// Traverse the array
for(int i = 0; i < N - 1; i++)
{
for(int j = i + 1; j < N; j++)
{
// If co-prime
if (coprime(Integer.parseInt(arr.get(i)),
Integer.parseInt(arr.get(j))))
{
// Increment count
count = count + 1;
}
}
}
// Return count
return count;
}
// Function to count number
// of co-prime pairs
static void noOfCoPrimePairs(int N, int d1, int d2)
{
// Stores digits in string form
ArrayList l = new ArrayList();
l.add(Integer.toString(d1));
l.add(Integer.toString(d2));
// Sort the list
Collections.sort(l);
if (N < Integer.parseInt(l.get(1)))
return;
// Keep two copies of list l
ArrayList total = new ArrayList(l);
ArrayList temp2 = new ArrayList(l);
int flag = 0;
ArrayList temp3 = new ArrayList(l);
// Generate 2 digit numbers
// using d1 and d2
while (l.get(0).length() < 10)
{
for(int i = 0; i < l.size(); i++)
{
for(int j = 0; j < 2; j++)
{
// If current number
// does not exceed N
if (Integer.parseInt(l.get(i) +
temp2.get(j)) > N)
{
flag = 1;
break;
}
total.add(l.get(i) + temp2.get(j));
temp3.add(l.get(i) + temp2.get(j));
}
if (flag == 1)
break;
}
if (flag == 1)
break;
l = temp3;
temp3.clear();
}
// Stores length of list
int lenOfTotal = total.size();
// Stores number of co-prime pairs
int ans = numOfPairs(total, lenOfTotal);
// Print number of co-prime pairs
System.out.print(ans);
}
// Driver Code
public static void main(String args[])
{
// Given value of N, d1, d2
int N = 30, d1 = 2, d2 = 3;
// Function call to count
// number of co-prime pairs
noOfCoPrimePairs(N, d1, d2);
}
}
// This code is contributed by bgangwar59 Python3
# Python3 program for the above approach
from copy import deepcopy
import math
# Function to check whether given
# integers are co-prime or not
def coprime(a, b):
return (math.gcd(a, b)) == 1
# Utility function to count
# number of co-prime pairs
def numOfPairs(arr, N):
count = 0
# Traverse the array
for i in range(0, N-1):
for j in range(i+1, N):
# If co-prime
if (coprime(int(arr[i]), int(arr[j]))):
# Increment count
count = count + 1
# Return count
return count
# Function to count number
# of co-prime pairs
def noOfCoPrimePairs(N, d1, d2):
# Stores digits in string form
l = []
l.append(str(d1))
l.append(str(d2))
# Sort the list
l.sort()
if int(N) < int(l[1]):
return 0
# Keep two copies of list l
total = temp2 = deepcopy(l)
flag = 0
temp3 = []
# Generate 2 digit numbers
# using d1 and d2
while len(l[0]) < 10:
for i in range(len(l)):
for j in range(2):
# If current number
# does not exceed N
if int(l[i]+temp2[j]) > int(N):
flag = 1
break
total.append(l[i]+temp2[j])
temp3.append(l[i]+temp2[j])
if flag == 1:
break
if flag == 1:
break
l = deepcopy(temp3)
temp3 = []
# Stores length of list
lenOfTotal = len(total)
# Stores number of co-prime pairs
ans = numOfPairs(total, lenOfTotal)
# Print number of co-prime pairs
print(ans)
# Driver Code
if __name__ == "__main__":
# Given value of N, d1, d2
N = 30
d1 = 2
d2 = 3
# Function call to count
# number of co-prime pairs
noOfCoPrimePairs(N, d1, d2)C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
static int GCD(int a, int b)
{
return b == 0 ? a : GCD(b, a % b);
}
// Function to check whether given
// integers are co-prime or not
static bool coprime(int a, int b)
{
if (GCD(a, b) == 1)
return true;
return false;
}
// Utility function to count
// number of co-prime pairs
static int numOfPairs(List arr, int N)
{
int count = 0;
// Traverse the array
for(int i = 0; i < N - 1; i++)
{
for(int j = i + 1; j < N; j++)
{
// If co-prime
if (coprime(Int32.Parse(arr[i]),
Int32.Parse(arr[j])))
{
// Increment count
count = count + 1;
}
}
}
// Return count
return count;
}
// Function to count number
// of co-prime pairs
static void noOfCoPrimePairs(int N, int d1, int d2)
{
// Stores digits in string form
List l = new List();
l.Add(d1.ToString());
l.Add(d2.ToString());
// Sort the list
l.Sort();
if (N < Int32.Parse(l[1]))
return;
// Keep two copies of list l
List total = new List(l);
List temp2 = new List(l);
int flag = 0;
List temp3 = new List();
// Generate 2 digit numbers
// using d1 and d2
while (l[0].Length < 10)
{
for(int i = 0; i < l.Count; i++)
{
for(int j = 0; j < 2; j++)
{
// If current number
// does not exceed N
if (Int32.Parse(l[i] + temp2[j]) > N)
{
flag = 1;
break;
}
total.Add(l[i] + temp2[j]);
temp3.Add(l[i] + temp2[j]);
}
if (flag == 1)
break;
}
if (flag == 1)
break;
l = temp3;
temp3.Clear();
}
// Stores length of list
int lenOfTotal = total.Count;
// Stores number of co-prime pairs
int ans = numOfPairs(total, lenOfTotal);
// Print number of co-prime pairs
Console.WriteLine(ans);
}
// Driver Code
public static void Main()
{
// Given value of N, d1, d2
int N = 30, d1 = 2, d2 = 3;
// Function call to count
// number of co-prime pairs
noOfCoPrimePairs(N, d1, d2);
}
}
// This code is contributed by ipg2016107 输出：
5时间复杂度： O(2 logN )
辅助空间： O(2 logN )