[P, Q] 范围内的对数为 R 的倍数，它们的乘积在 [P*Q/4, P*Q] 范围内
给定3 个正整数P 、 Q和R ，任务是找到对的数量，使得两个元素都在[P, Q]范围内，并且数字应该是R的倍数，并且数字的乘积应该位于在[P × Q / 4, P × Q]范围内。如果不存在这样的对，则打印-1。
例子：
Input: P = 14, Q = 30, R = 5
Output:15 20
15 25
Explanation:
Multiple of R between P & Q are {15, 20, 25, 30}.
P × Q = 420 and P × Q / 4 = 105
So the pairs which satisfies the above conditions are 15, 20 and 15, 25.
Input: P = 10, Q = 20, R = 7
Output: 7 14
方法：为了解决这个问题，首先找到可以存在的对的最小和最大范围，然后找到满足上述条件的对。请按照以下步骤解决问题：
- 初始化向量说， v来存储所有在[P, Q]范围内并且是R的倍数的数和一个对向量，比如ans来存储满足上述条件的对。
- 使用变量i在[P, Q]范围内迭代并检查i是否可被R整除，然后将i插入向量v中。
- 使用变量i在[0, v.size()-1]范围内迭代并执行以下步骤：
- 使用变量j在[i+1, v.size()-1]范围内迭代并检查是否v[j] * v[i] <= P * Q和v[j] * v[i] >= P * Q/4然后将该对插入ans中。
- 如果ans.size()等于0 ，则打印-1。
- 否则，打印ans中的对。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the number of
// pairs such that both the elements
// are in the range [P, Q] and the
// numbers should be multiple of R,
// and the product of numbers should
// lie in the range [P*Q/4, P*Q]
void findPairs(int p, int q, int r)
{
// Store multiple of r
// in range of [P, Q]
vector v;
// Iterate in the range [p, q]
for (int i = p; i <= q; i++) {
if (i % r == 0) {
v.push_back(i);
}
}
// Vector to store pair of answer
vector > ans;
// Iterate through the vector v
for (int i = 0; i < v.size(); i++) {
// Iterate in the range [i+1, v.size()-1]
for (int j = i + 1; j < v.size(); j++) {
// If pair follow this condition
// insert the pair in vector ans
if (v[i] * v[j] >= p * q / 4
&& v[i] * v[j] <= p * q) {
ans.push_back({ v[i], v[j] });
}
}
}
// If no pair satisfy the conditions, print -1
if (ans.size() == 0) {
cout << -1 << endl;
}
else {
// Print the pairs
// which satisfy the given condition
for (int i = 0; i < ans.size(); i++) {
cout << ans[i].first << " "
<< ans[i].second << endl;
}
}
}
// Driver Code
int main()
{
// Given Input
int p = 14, q = 30, r = 5;
// Function Call
findPairs(p, q, r);
return 0;
} Java
//Java program for above approach
import java.awt.*;
import java.util.*;
class GFG{
static class pair< T, V>{
T first;
V second;
}
// Function to find the number of
// pairs such that both the elements
// are in the range [P, Q] and the
// numbers should be multiple of R,
// and the product of numbers should
// lie in the range [P*Q/4, P*Q]
static void findPairs(int p, int q, int r)
{
// Store multiple of r
// in range of [P, Q]
ArrayList v = new ArrayList<>();
// Iterate in the range [p, q]
for (int i = p; i <= q; i++) {
if (i % r == 0) {
v.add(i);
}
}
// Vector to store pair of answer
ArrayList > ans = new ArrayList<>();
// Iterate through the vector v
for (int i = 0; i < v.size(); i++) {
// Iterate in the range [i+1, v.size()-1]
for (int j = i + 1; j < v.size(); j++) {
// If pair follow this condition
// insert the pair in vector ans
if (v.get(i) * v.get(j) >= p * q / 4
&& v.get(i) * v.get(j) <= p * q) {
pair x = new pair<>();
x.first = v.get(i);
x.second = v.get(j);
ans.add(x);
}
}
}
// If no pair satisfy the conditions, print -1
if (ans.size() == 0) {
System.out.println(-1);
}
else {
// Print the pairs
// which satisfy the given condition
for (int i = 0; i < ans.size(); i++) {
System.out.println(ans.get(i).first +
" " + ans.get(i).second);
}
}
}
// Driver Code
public static void main(String[] args)
{
// Given Input
int p = 14, q = 30, r = 5;
// Function Call
findPairs(p, q, r);
}
}
// This code is contributed by hritikrommie. Python3
# Python3 program for the above approach
# Function to find the number of
# pairs such that both the elements
# are in the range [P, Q] and the
# numbers should be multiple of R,
# and the product of numbers should
# lie in the range [P*Q/4, P*Q]
def findPairs(p, q, r):
# Store multiple of r
# in range of [P, Q]
v = []
# Iterate in the range [p, q]
for i in range(p, q + 1):
if (i % r == 0):
v.append(i)
# Vector to store pair of answer
ans = []
# Iterate through the vector v
for i in range(len(v)):
# Iterate in the range [i+1, v.size()-1]
for j in range(i + 1, len(v)):
# If pair follow this condition
# insert the pair in vector ans
if (v[i] * v[j] >= p * q // 4 and
v[i] * v[j] <= p * q):
ans.append([v[i], v[j]])
# If no pair satisfy the conditions, pr-1
if (len(ans) == 0):
print (-1)
else:
# Print the pairs
# which satisfy the given condition
for i in range(len(ans)):
print(ans[i][0], ans[i][1])
# Driver Code
if __name__ == '__main__':
# Given Input
p = 14
q = 30
r = 5
# Function Call
findPairs(p, q, r)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the number of
// pairs such that both the elements
// are in the range [P, Q] and the
// numbers should be multiple of R,
// and the product of numbers should
// lie in the range [P*Q/4, P*Q]
static void findPairs(int p, int q, int r)
{
// Store multiple of r
// in range of [P, Q]
List v = new List();
// Iterate in the range [p, q]
for (int i = p; i <= q; i++) {
if (i % r == 0) {
v.Add(i);
}
}
// Vector to store pair of answer
List> ans = new List>();
// Iterate through the vector v
for(int i = 0; i < v.Count; i++) {
// Iterate in the range [i+1, v.size()-1]
for (int j = i + 1; j < v.Count; j++) {
// If pair follow this condition
// insert the pair in vector ans
if (v[i] * v[j] >= p * q / 4
&& v[i] * v[j] <= p * q) {
List temp = new List();
temp.Add(v[i]);
temp.Add(v[j]);
ans.Add(temp);
}
}
}
// If no pair satisfy the conditions, print -1
if (ans.Count == 0) {
Console.Write(-1);
}
else {
foreach (List subList in ans)
{
foreach (int item in subList)
{
Console.Write(item + " ");
}
Console.WriteLine();
}
// Print the pairs
// which satisfy the given condition
}
}
// Driver Code
public static void Main()
{
// Given Input
int p = 14, q = 30, r = 5;
// Function Call
findPairs(p, q, r);
}
}
// This code is contributed by ipg2016107. Javascript
输出：
15 20
15 25时间复杂度： O(N 2 )，其中N是Q – P + 1。
辅助空间： O(N)