给定的一组数字，发现将L ongest在它ģeometrix P rogression(LLGP)的L个ength。 GP的公共比率必须为整数。
例子：
set[] = {5, 7, 10, 15, 20, 29}
output = 3
The longest geometric progression is {5, 10, 20}
set[] = {3, 9, 27, 81}
output = 4
此问题类似于最长算术级数问题。我们可以使用动态编程解决此问题。
我们首先对给定的集合进行排序。我们使用辅助表L [n] [n]存储子问题的结果。该表中的条目L [i] [j]存储LLGP，其中set [i]和set [j]作为GP的前两个元素，且j> i。该表从右下到左上填充。为了填充表格，首先固定j(GP中的第二个元素)。在i和k中搜索固定的j。如果找到i和k使得i，j，k形成GP，则将L [i] [j]的值设置为L [j] [k] +1。请注意，L [j]的值[k]必须先填充，然后循环才能从右到左遍历。
以下是动态编程算法的实现。
C++
// C++ program to find length
// of the longest geometric
// progression in a given set
#include
#include
using namespace std;
// Returns length of the
// longest GP subset of set[]
int lenOfLongestGP(int set[], int n)
{
// Base cases
if (n < 2)
return n;
if (n == 2)
return (set[1] % set[0] == 0) ? 2 : 1;
// Let us sort the set first
sort(set, set+n);
// An entry L[i][j] in this
// table stores LLGP with
// set[i] and set[j] as first
// two elements of GP
// and j > i.
int L[n][n];
// Initialize result (A single element
// is always a GP)
int llgp = 1;
// Initialize values of last column
for (int i = 0; i < n - 1; ++i) {
if (set[n-1] % set[i] == 0)
{
L[i][n-1] = 2;
if (2 > llgp)
llgp = 2;
}
else
{
L[i][n-1] = 1;
}
}
L[n-1][n-1] = 1;
// Consider every element as
// second element of GP
for (int j = n - 2; j >= 1; --j)
{
// Search for i and k for j
int i = j - 1, k = j+1;
while (i>=0 && k <= n-1)
{
// Two cases when i, j and k don't form
// a GP.
if (set[i] * set[k] < set[j]*set[j])
{
++k;
}
else if (set[i] * set[k] > set[j]*set[j])
{
if (set[j] % set[i] == 0)
{
L[i][j] = 2;
}
else
{
L[i][j] = 1;
}
--i;
}
// i, j and k form GP, LLGP with i and j as
// first two elements is equal to LLGP with
// j and k as first two elements plus 1.
// L[j][k] must have been filled before as
// we run the loop from right side
else
{
if (set[j] % set[i] == 0)
{
L[i][j] = L[j][k] + 1;
// Update overall LLGP
if (L[i][j] > llgp)
llgp = L[i][j];
} else {
L[i][j] = 1;
}
// Change i and k to fill more L[i][j]
// values for current j
--i;
++k;
}
}
// If the loop was stopped due to k becoming
// more than n-1, set the remaining entries
// in column j as 1 or 2 based on divisibility
// of set[j] by set[i]
while (i >= 0)
{
if (set[j] % set[i] == 0)
{
L[i][j] = 2;
if (2 > llgp)
llgp = 2;
}
else
L[i][j] = 1;
--i;
}
}
// Return result
return llgp;
}
// Driver code
int main()
{
int set1[] = {1, 3, 9, 27, 81, 243};
int n1 = sizeof(set1)/sizeof(set1[0]);
cout << lenOfLongestGP(set1, n1) << "\n";
int set2[] = {1, 3, 4, 9, 7, 27};
int n2 = sizeof(set2)/sizeof(set2[0]);
cout << lenOfLongestGP(set2, n2) << "\n";
int set3[] = {2, 3, 5, 7, 11, 13};
int n3 = sizeof(set3)/sizeof(set3[0]);
cout << lenOfLongestGP(set3, n3) << "\n";
return 0;
} Java
// Java program to find length
// of the longest geometric
// progression in a given set
import java.util.*;
class GFG {
// Returns length of the longest GP subset of set[]
static int lenOfLongestGP(int set[], int n)
{
// Base cases
if (n < 2) {
return n;
}
if (n == 2) {
return (set[1] % set[0] == 0 ? 2 : 1);
}
// Let us sort the set first
Arrays.sort(set);
// An entry L[i][j] in this table
// stores LLGP with set[i] and set[j]
// as first two elements of GP
// and j > i.
int L[][] = new int[n][n];
// Initialize result (A single
// element is always a GP)
int llgp = 1;
// Initialize values of last column
for (int i = 0; i < n - 1; ++i) {
if (set[n - 1] % set[i] == 0) {
L[i][n - 1] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][n - 1] = 1;
}
}
L[n - 1][n - 1] = 1;
// Consider every element as second element of GP
for (int j = n - 2; j >= 1; --j) {
// Search for i and k for j
int i = j - 1, k = j + 1;
while (i >= 0 && k <= n - 1) {
// Two cases when i, j and k
// don't form a GP.
if (set[i] * set[k] < set[j] * set[j]) {
++k;
}
else if (set[i] * set[k]
> set[j] * set[j]) {
if (set[j] % set[i] == 0) {
L[i][j] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][j] = 1;
}
--i;
}
// i, j and k form GP, LLGP with i and j as
// first two elements is equal to LLGP with
// j and k as first two elements plus 1.
// L[j][k] must have been filled before as
// we run the loop from right side
else {
if (set[j] % set[i] == 0) {
L[i][j] = L[j][k] + 1;
// Update overall LLGP
if (L[i][j] > llgp) {
llgp = L[i][j];
}
}
else {
L[i][j] = 1;
}
// Change i and k to fill more L[i][j]
// values for current j
--i;
++k;
}
}
// If the loop was stopped due to k becoming
// more than n-1, set the remaining entries
// in column j as 1 or 2 based on divisibility
// of set[j] by set[i]
while (i >= 0) {
if (set[j] % set[i] == 0) {
L[i][j] = 2;
if (2 > llgp)
llgp = 2;
}
else {
L[i][j] = 1;
}
--i;
}
}
// Return result
return llgp;
}
// Driver code
public static void main(String[] args)
{
int set1[] = { 1, 3, 9, 27, 81, 243 };
int n1 = set1.length;
System.out.print(lenOfLongestGP(set1, n1) + "\n");
int set2[] = { 1, 3, 4, 9, 7, 27 };
int n2 = set2.length;
System.out.print(lenOfLongestGP(set2, n2) + "\n");
int set3[] = { 2, 3, 5, 7, 11, 13 };
int n3 = set3.length;
System.out.print(lenOfLongestGP(set3, n3) + "\n");
}
}
/* This code contributed by PrinciRaj1992 */Python3
# Python3 program to find length
# of the longest geometric
# progression in a given sett
# Returns length of the longest GP
# subset of sett[]
def lenOfLongestGP(sett, n):
# Base cases
if n < 2:
return n
if n == 2:
return 2 if (sett[1] % sett[0] == 0) else 1
# let us sort the sett first
sett.sort()
# An entry L[i][j] in this
# table stores LLGP with
# sett[i] and sett[j] as first
# two elements of GP
# and j > i.
L = [[0 for i in range(n)] for i in range(n)]
# Initialize result (A single
# element is always a GP)
llgp = 1
# Initialize values of last column
for i in range(0, n-1):
if sett[n-1] % sett[i] == 0:
L[i][n-1] = 2
if 2 > llgp:
llgp = 2
else:
L[i][n-1] = 1
L[n-1][n-1] = 1
# Consider every element as second element of GP
for j in range(n-2, 0, -1):
# Search for i and k for j
i = j - 1
k = j + 1
while i >= 0 and k <= n - 1:
# Two cases when i, j and k don't form
# a GP.
if sett[i] * sett[k] < sett[j] * sett[j]:
k += 1
elif sett[i] * sett[k] > sett[j] * sett[j]:
if sett[j] % sett[i] == 0:
L[i][j] = 2
else:
L[i][j] = 1
i -= 1
# i, j and k form GP, LLGP with i and j as
# first two elements is equal to LLGP with
# j and k as first two elements plus 1.
# L[j][k] must have been filled before as
# we run the loop from right side
else:
if sett[j] % sett[i] == 0:
L[i][j] = L[j][k] + 1
# Update overall LLGP
if L[i][j] > llgp:
llgp = L[i][j]
else:
L[i][j] = 1
# Change i and k to fill more L[i][j]
# values for current j
i -= 1
k += 1
# If the loop was stopped due to k becoming
# more than n-1, set the remaining entries
# in column j as 1 or 2 based on divisibility
# of sett[j] by sett[i]
while i >= 0:
if sett[j] % sett[i] == 0:
L[i][j] = 2
else:
L[i][j] = 1
i -= 1
return llgp
# Driver code
if __name__ == '__main__':
set1 = [1, 3, 9, 27, 81, 243]
n1 = len(set1)
print(lenOfLongestGP(set1, n1))
set2 = [1, 3, 4, 9, 7, 27]
n2 = len(set2)
print(lenOfLongestGP(set2, n2))
set3 = [2, 3, 5, 7, 11, 13]
n3 = len(set3)
print(lenOfLongestGP(set3, n3))
# this code is contrubuted by sahilshelangiaC#
// C# program to find length
// of the longest geometric
// progression in a given Set
using System;
class GFG
{
// Returns length of the
// longest GP subset of Set[]
static int lenOfLongestGP(int []Set, int n)
{
// Base cases
if (n < 2)
{
return n;
}
if (n == 2)
{
return (Set[1] % Set[0] == 0 ? 2 : 1);
}
// Let us sort the Set first
Array.Sort(Set);
// An entry L[i,j] in this table
// stores LLGP with Set[i] and Set[j]
// as first two elements of GP
// and j > i.
int [,]L = new int[n, n];
// Initialize result (A single
// element is always a GP)
int llgp = 1;
// Initialize values of last column
for (int i = 0; i < n - 1; ++i)
{
if (Set[n - 1] % Set[i] == 0)
{
L[i, n - 1] = 2;
if (2 > llgp)
llgp = 2;
}
else
{
L[i, n - 1] = 1;
}
}
L[n - 1, n - 1] = 1;
// Consider every element
// as second element of GP
for (int j = n - 2; j >= 1; --j)
{
// Search for i and k for j
int i = j - 1, k = j + 1;
while (i >= 0 && k <= n - 1)
{
// Two cases when i, j and k
// don't form a GP.
if (Set[i] * Set[k] < Set[j] * Set[j])
{
++k;
}
else if (Set[i] * Set[k] > Set[j] * Set[j])
{
if (Set[j] % Set[i] == 0)
{
L[i,j] = 2;
if (2 > llgp)
llgp = 2;
}
else
{
L[i,j] = 1;
}
--i;
}
// i, j and k form GP, LLGP with i and j as
// first two elements is equal to LLGP with
// j and k as first two elements plus 1.
// L[j,k] must have been filled before as
// we run the loop from right side
else
{
if (Set[j] % Set[i] == 0)
{
L[i, j] = L[j, k] + 1;
// Update overall LLGP
if (L[i, j] > llgp)
{
llgp = L[i, j];
}
}
else
{
L[i, j] = 1;
}
// Change i and k to fill more L[i,j]
// values for current j
--i;
++k;
}
}
// If the loop was stopped due to k becoming
// more than n-1, set the remaining entries
// in column j as 1 or 2 based on divisibility
// of Set[j] by Set[i]
while (i >= 0)
{
if (Set[j] % Set[i] == 0)
{
L[i, j] = 2;
if (2 > llgp)
llgp = 2;
}
else
{
L[i, j] = 1;
}
--i;
}
}
// Return result
return llgp;
}
// Driver code
public static void Main(String[] args)
{
int []set1 = {1, 3, 9, 27, 81, 243};
int n1 = set1.Length;
Console.Write(lenOfLongestGP(set1, n1) + "\n");
int []set2 = {1, 3, 4, 9, 7, 27};
int n2 = set2.Length;
Console.Write(lenOfLongestGP(set2, n2) + "\n");
int []set3 = {2, 3, 5, 7, 11, 13};
int n3 = set3.Length;
Console.Write(lenOfLongestGP(set3, n3) + "\n");
}
}
// This code has been contributed by 29AjayKumar输出：
6
4
1