Dada una array A de n enteros y dos enteros p y q, la tarea es aumentar algunos (o todos) los valores de la array A de tal manera que la relación de cada elemento (después del primer elemento) con respecto a la suma total de todos los elementos antes de que el elemento actual permanezca menor o igual a p / q . Devuelve la suma mínima de cambios a realizar.
Ejemplos:
Entrada : n = 4, A = {20100, 1, 202, 202}, p = 1, q = 100
Salida : 99
Explicación : 50 se agrega al primer elemento y 49 al segundo elemento, de modo que la array resultante se convierte en {20150, 50, 202, 202}.
50/20150 <= 1/100
202/(20150+50) <=1/100
202/(20150+50+202)<=1/100
Por lo tanto, la condición se cumple para todos los elementos y por lo tanto 50+49= 99 sería la respuesta.
También hay otras respuestas posibles, pero necesitamos encontrar la suma mínima posible.Entrada : n = 3, A = {1, 1, 1}, p = 100, q = 100
Salida : 0
Enfoque: El problema se puede resolver fácilmente utilizando los conceptos de suma de prefijos y búsqueda binaria .
- Se puede observar claramente que
- si la condición establecida en el problema se logra para una cierta suma de cambios (digamos S ),
- entonces siempre es posible lograr la condición para todos los números mayores que S y
- no se puede lograr para todos los números menores que S.
- Entonces, podemos aplicar la búsqueda binaria para encontrar la respuesta.
- Además, se debe observar cuidadosamente que en lugar de distribuir S (suma de cambios) entre diferentes elementos, si solo lo agregamos al primer elemento, eso no afectaría la respuesta.
Por ejemplo, en el primer ejemplo anterior, si se agrega 99 al primer elemento, la array resultante aún cumplirá la condición requerida.
Siga los pasos a continuación para resolver el problema:
- Haga una array de suma de prefijos de la array dada.
- Usando la búsqueda binaria en el rango 0 a INT_MAX, encuentre la respuesta mínima posible.
NOTA: La división puede generar errores y valores superpuestos.
Entonces, en lugar de una comparación como (a/b)<=(c/d) , haremos (a*d)<=(b*c) .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for find minimum
// sum of changes in an array
#include <bits/stdc++.h>
using namespace std;
// function to check if the candidate sum
// satisfies the condition of the problem
bool isValid(int candidate, int pre[], int n, int A[],
int p, int q)
{
// flag variable to check wrong answer
bool flag = true;
for (int i = 1; i < n; i++) {
// Now for each element, we are checking
// if its ratio with sum of all previous
// elements + candidate is greater than p/q.
// If so, we will return false.
int curr_sum = pre[i - 1] + candidate;
if (A[i] * q > p * curr_sum) {
flag = false;
break;
}
// comparing like A[i]/(curr_sum)>p/q
// will be error prone.
}
return flag;
}
int solve(int n, int A[], int p, int q)
{
// declaring and constructing
// prefix sum array
int pre[n];
pre[0] = A[0];
for (int i = 1; i < n; i++) {
pre[i] = A[i] + pre[i - 1];
}
// setting lower and upper bound for
// binary search
int lo = 0, hi = INT_MAX, ans = INT_MAX;
// since minimum answer is needed,
// so it is initialized with INT_MAX
while (lo <= hi) {
// calculating mid by using (lo+hi)/2
// may overflow in certain cases
int mid = lo + (hi - lo) / 2;
// checking if required ratio would be
// achieved by all elements if "mid" is
// considered as answer
if (isValid(mid, pre, n, A, p, q)) {
ans = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return ans;
}
// Driver Function
int main()
{
int n, p, q;
n = 4, p = 1, q = 100;
int A[] = { 20100, 1, 202, 202 };
// printing the required answer
cout << solve(n, A, p, q) << endl;
}
Java
// Java program for find minimum
// sum of changes in an arraykage whatever //do not write package name here */
import java.io.*;
class GFG {
// function to check if the candidate sum
// satisfies the condition of the problem
static Boolean isValid(int candidate, int pre[], int n, int A[],
int p, int q)
{
// flag variable to check wrong answer
Boolean flag = true;
for (int i = 1; i < n; i++) {
// Now for each element, we are checking
// if its ratio with sum of all previous
// elements + candidate is greater than p/q.
// If so, we will return false.
int curr_sum = pre[i - 1] + candidate;
if (A[i] * q > p * curr_sum) {
flag = false;
break;
}
// comparing like A[i]/(curr_sum)>p/q
// will be error prone.
}
return flag;
}
static int solve(int n, int A[], int p, int q)
{
// declaring and constructing
// prefix sum array
int pre[] = new int[n];
pre[0] = A[0];
for (int i = 1; i < n; i++) {
pre[i] = A[i] + pre[i - 1];
}
// setting lower and upper bound for
// binary search
int lo = 0, hi = Integer.MAX_VALUE, ans = Integer.MAX_VALUE;
// since minimum answer is needed,
// so it is initialized with INT_MAX
while (lo <= hi) {
// calculating mid by using (lo+hi)/2
// may overflow in certain cases
int mid = lo + (hi - lo) / 2;
// checking if required ratio would be
// achieved by all elements if "mid" is
// considered as answer
if (isValid(mid, pre, n, A, p, q)) {
ans = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return ans;
}
// Driver Function
public static void main (String[] args)
{
int n = 4, p = 1, q = 100;
int A[] = { 20100, 1, 202, 202 };
// printing the required answer
System.out.println(solve(n, A, p, q));
}
}
// This code is contributed by hrithikgarg03188.
Python3
# Python code for the above approach import sys # function to check if the candidate sum # satisfies the condition of the problem def isValid(candidate, pre, n, A, p, q) : # flag variable to check wrong answer flag = True for i in range(1, n) : # Now for each element, we are checking # if its ratio with sum of all previous # elements + candidate is greater than p/q. # If so, we will return false. curr_sum = pre[i - 1] + candidate if (A[i] * q > p * curr_sum) : flag = False break # comparing like A[i]/(curr_sum)>p/q # will be error prone. return flag def solve(n, A, p, q) : # declaring and constructing # prefix sum array pre = [0] * 100 pre[0] = A[0] for i in range(1, n) : pre[i] = A[i] + pre[i - 1] # setting lower and upper bound for # binary search lo = 0 hi = sys.maxsize ans = sys.maxsize # since minimum answer is needed, # so it is initialized with INT_MAX while (lo <= hi) : # calculating mid by using (lo+hi)/2 # may overflow in certain cases mid = lo + (hi - lo) // 2 # checking if required ratio would be # achieved by all elements if "mid" is # considered as answer if (isValid(mid, pre, n, A, p, q)) : ans = mid hi = mid - 1 else : lo = mid + 1 return ans # Driver Function n = 4 p = 1 q = 100 A = [ 20100, 1, 202, 202 ] # printing the required answer print(solve(n, A, p, q)) # This code is contributed by code_hunt.
C#
// C# program for find minimum
// sum of changes in an array
using System;
class GFG
{
// function to check if the candidate sum
// satisfies the condition of the problem
static bool isValid(int candidate, int []pre, int n, int []A,
int p, int q)
{
// flag variable to check wrong answer
bool flag = true;
for (int i = 1; i < n; i++) {
// Now for each element, we are checking
// if its ratio with sum of all previous
// elements + candidate is greater than p/q.
// If so, we will return false.
int curr_sum = pre[i - 1] + candidate;
if (A[i] * q > p * curr_sum) {
flag = false;
break;
}
// comparing like A[i]/(curr_sum)>p/q
// will be error prone.
}
return flag;
}
static int solve(int n, int []A, int p, int q)
{
// declaring and constructing
// prefix sum array
int []pre = new int[n];
pre[0] = A[0];
for (int i = 1; i < n; i++) {
pre[i] = A[i] + pre[i - 1];
}
// setting lower and upper bound for
// binary search
int lo = 0, hi = Int32.MaxValue, ans = Int32.MaxValue;
// since minimum answer is needed,
// so it is initialized with INT_MAX
while (lo <= hi) {
// calculating mid by using (lo+hi)/2
// may overflow in certain cases
int mid = lo + (hi - lo) / 2;
// checking if required ratio would be
// achieved by all elements if "mid" is
// considered as answer
if (isValid(mid, pre, n, A, p, q)) {
ans = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return ans;
}
// Driver Function
public static void Main()
{
int n = 4, p = 1, q = 100;
int []A = { 20100, 1, 202, 202 };
// printing the required answer
Console.WriteLine(solve(n, A, p, q));
}
}
// This code is contributed by Samim Hossain Mondal.
Javascript
<script>
// JavaScript program for find minimum
// sum of changes in an array
const INT_MAX = 2147483647;
// function to check if the candidate sum
// satisfies the condition of the problem
const isValid = (candidate, pre, n, A, p, q) => {
// flag variable to check wrong answer
let flag = true;
for (let i = 1; i < n; i++) {
// Now for each element, we are checking
// if its ratio with sum of all previous
// elements + candidate is greater than p/q.
// If so, we will return false.
let curr_sum = pre[i - 1] + candidate;
if (A[i] * q > p * curr_sum) {
flag = false;
break;
}
// comparing like A[i]/(curr_sum)>p/q
// will be error prone.
}
return flag;
}
let solve = (n, A, p, q) => {
// declaring and constructing
// prefix sum array
let pre = new Array(n).fill(0);
pre[0] = A[0];
for (let i = 1; i < n; i++) {
pre[i] = A[i] + pre[i - 1];
}
// setting lower and upper bound for
// binary search
let lo = 0, hi = INT_MAX, ans = INT_MAX;
// since minimum answer is needed,
// so it is initialized with INT_MAX
while (lo <= hi) {
// calculating mid by using (lo+hi)/2
// may overflow in certain cases
let mid = lo + parseInt((hi - lo) / 2);
// checking if required ratio would be
// achieved by all elements if "mid" is
// considered as answer
if (isValid(mid, pre, n, A, p, q)) {
ans = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return ans;
}
// Driver Function
let n, p, q;
n = 4, p = 1, q = 100;
let A = [20100, 1, 202, 202];
// printing the required answer
document.write(solve(n, A, p, q));
// This code is contributed by rakeshsahni
</script>
Producción
99
Complejidad de tiempo : O(n log(INT_MAX))
Espacio auxiliar : O(n)
Publicación traducida automáticamente
Artículo escrito por kamabokogonpachiro y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA