Минимальное количество операций для преобразования заданной последовательности в геометрическую прогрессию
Учитывая последовательность из N элементов, только три операции могут быть выполнены с любым элементом не более одного раза. Операции следующие:
- Добавьте один к элементу.
- Вычтите единицу из элемента.
- Оставьте элемент без изменений.
Выполните любую из операций над всеми элементами массива. Задача состоит в том, чтобы найти минимальное количество операций (сложение и вычитание), которые можно выполнить с последовательностью, чтобы преобразовать ее в геометрическую прогрессию. Если невозможно сгенерировать GP, выполнив указанные выше операции, выведите -1.
Примеры :
Input: a[] = {1, 1, 4, 7, 15, 33}
Output: The minimum number of operations are 4.
Steps:
- Keep a1 unchanged
- Add one to a2.
- Keep a3 unchanged
- Subtract one from a4.
- Subtract one from a5.
- Add one to a6.
The resultant sequence is {1, 2, 4, 8, 16, 32}
Input: a[] = {20, 15, 20, 15}
Output: -1
Подход Ключевое наблюдение, которое следует сделать здесь, заключается в том, что любая геометрическая прогрессия однозначно определяется только своими первыми двумя элементами (поскольку соотношение между каждой из следующих пар должно быть таким же, как соотношение между этой парой, состоящей из первых двух элементов ). Поскольку возможны только 3 * 3 перестановки. Возможная комбинация операций: (+1, +1), (+1, 0), (+1, -1), (-1, +1), (-1, 0), (-1, -1. ), (0, +1), (0, 0) и (0, -1). Использование грубой силы всех этих 9 перестановок и проверка того, образуют ли они GP за линейное время, даст нам ответ. Ответом будет минимум операций, которые приводят к комбинациям, которые находятся в GP.
Below is the implementation of the above approach:
C++
// C++ program to find minimum number // of operations to convert a given // sequence to an Geometric Progression #include <bits/stdc++.h> using namespace std; // Function to print the GP series void construct( int n, pair< double , double > ans_pair) { // Check for possibility if (ans_pair.first == -1) { cout << "Not possible" ; return ; } double a1 = ans_pair.first; double a2 = ans_pair.second; double r = a2 / a1; cout << "The resultant sequence is:
" ; for ( int i = 1; i <= n; i++) { double ai = a1 * pow (r, i - 1); cout << ai << " " ; } } // Function for getting the Arithmetic Progression void findMinimumOperations( double * a, int n) { int ans = INT_MAX; // The array c describes all the given set of // possible operations. int c[] = { -1, 0, 1 }; // Size of c int possiblities = 3; // candidate answer int pos1 = -1, pos2 = -1; // loop through all the permutations of the first two // elements. for ( int i = 0; i < possiblities; i++) { for ( int j = 0; j < possiblities; j++) { // a1 and a2 are the candidate first two elements // of the possible GP. double a1 = a[1] + c[i]; double a2 = a[2] + c[j]; // temp stores the current answer, including the // modification of the first two elements. int temp = abs (a1 - a[1]) + abs (a2 - a[2]); if (a1 == 0 || a2 == 0) continue ; // common ratio of the possible GP double r = a2 / a1; // To check if the chosen set is valid, and id yes // find the number of operations it takes. for ( int pos = 3; pos <= n; pos++) { // ai is value of a[i] according to the assumed // first two elements a1, a2 // ith element of an GP = a1*((a2-a1)^(i-1)) double ai = a1 * pow (r, pos - 1); // Check for the "proposed" element to be only // differing by one if (a[pos] == ai) { continue ; } else if (a[pos] + 1 == ai || a[pos] - 1 == ai) { temp++; } else { temp = INT_MAX; // set the temporary ans break ; // to infinity and break } } // update answer if (temp < ans) { ans = temp; pos1 = a1; pos2 = a2; } } } if (ans == -1) { cout << "-1" ; return ; } cout << "Minimum Number of Operations are " << ans << "
" ; pair< double , double > ans_pair = { pos1, pos2 }; // Calling function to print the sequence construct(n, ans_pair); } // Driver Code int main() { // array is 1-indexed, with a[0] = 0 // for the sake of simplicity double a[] = { 0, 7, 20, 49, 125 }; int n = sizeof (a) / sizeof (a[0]); // Function to print the minimum operations // and the sequence of elements findMinimumOperations(a, n - 1); return 0; } |
Java
// Java program to find minimum number // of operations to convert a given // sequence to an Geometric Progression import java.util.*; class GFG { static class pair { double first, second; public pair( double first, double second) { this .first = first; this .second = second; } } // Function to print the GP series static void construct( int n, pair ans_pair) { // Check for possibility if (ans_pair.first == - 1 ) { System.out.print( "Not possible" ); return ; } double a1 = ans_pair.first; double a2 = ans_pair.second; double r = a2 / a1; System.out.print( "The resultant sequence is:
" ); for ( int i = 1 ; i <= n; i++) { int ai = ( int ) (a1 * Math.pow(r, i - 1 )); System.out.print(ai + " " ); } } // Function for getting the Arithmetic Progression static void findMinimumOperations( double []a, int n) { int ans = Integer.MAX_VALUE; // The array c describes all the given set of // possible operations. int c[] = { - 1 , 0 , 1 }; // Size of c int possiblities = 3 ; // candidate answer int pos1 = - 1 , pos2 = - 1 ; // loop through all the permutations of the first two // elements. for ( int i = 0 ; i < possiblities; i++) { for ( int j = 0 ; j < possiblities; j++) { // a1 and a2 are the candidate first two elements // of the possible GP. double a1 = a[ 1 ] + c[i]; double a2 = a[ 2 ] + c[j]; // temp stores the current answer, including the // modification of the first two elements. int temp = ( int ) (Math.abs(a1 - a[ 1 ]) + Math.abs(a2 - a[ 2 ])); if (a1 == 0 || a2 == 0 ) continue ; // common ratio of the possible GP double r = a2 / a1; // To check if the chosen set is valid, and id yes // find the number of operations it takes. for ( int pos = 3 ; pos <= n; pos++) { // ai is value of a[i] according to the assumed // first two elements a1, a2 // ith element of an GP = a1*((a2-a1)^(i-1)) double ai = a1 * Math.pow(r, pos - 1 ); // Check for the "proposed" element to be only // differing by one if (a[pos] == ai) { continue ; } else if (a[pos] + 1 == ai || a[pos] - 1 == ai) { temp++; } else { temp = Integer.MAX_VALUE; // set the temporary ans break ; // to infinity and break } } // update answer if (temp < ans) { ans = temp; pos1 = ( int ) a1; pos2 = ( int ) a2; } } } if (ans == - 1 ) { System.out.print( "-1" ); return ; } System.out.print( "Minimum Number of Operations are " + ans+ "
" ); pair ans_pair = new pair( pos1, pos2 ); // Calling function to print the sequence construct(n, ans_pair); } // Driver Code public static void main(String[] args) { // array is 1-indexed, with a[0] = 0 // for the sake of simplicity double a[] = { 0 , 7 , 20 , 49 , 125 }; int n = a.length; // Function to print the minimum operations // and the sequence of elements findMinimumOperations(a, n - 1 ); } } // This code is contributed by 29AjayKumar |
Python3
# Python program to find minimum number # of operations to convert a given # sequence to an Geometric Progression from sys import maxsize as INT_MAX # Function to print the GP series def construct(n: int , ans_pair: tuple ): # Check for possibility if ans_pair[ 0 ] = = - 1 : print ( "Not possible" ) return a1 = ans_pair[ 0 ] a2 = ans_pair[ 1 ] r = a2 / a1 print ( "The resultant sequence is" ) for i in range ( 1 , n + 1 ): ai = a1 * pow (r, i - 1 ) print ( int (ai), end = " " ) # Function for getting the Arithmetic Progression def findMinimumOperations(a: list , n: int ): ans = INT_MAX # The array c describes all the given set of # possible operations. c = [ - 1 , 0 , 1 ] # Size of c possibilities = 3 # candidate answer pos1 = - 1 pos2 = - 1 # loop through all the permutations of the first two # elements. for i in range (possibilities): for j in range (possibilities): # a1 and a2 are the candidate first two elements # of the possible GP. a1 = a[ 1 ] + c[i] a2 = a[ 2 ] + c[j] # temp stores the current answer, including the # modification of the first two elements. temp = abs (a1 - a[ 1 ]) + abs (a2 - a[ 2 ]) if a1 = = 0 or a2 = = 0 : continue # common ratio of the possible GP r = a2 / a1 # To check if the chosen set is valid, and id yes # find the number of operations it takes. for pos in range ( 3 , n + 1 ): # ai is value of a[i] according to the assumed # first two elements a1, a2 # ith element of an GP = a1*((a2-a1)^(i-1)) ai = a1 * pow (r, pos - 1 ) # Check for the "proposed" element to be only РЕКОМЕНДУЕМЫЕ СТАТЬИ |