数组的 GCD 大于一
给定一个包含 n 个整数的数组。如果所有元素的 GCD 都大于 1，则认为该数组是最好的。如果该数组不是最好的，我们可以选择索引 i (1 <= i < n) 并将数字 a i和 a i + 1替换为 a i – 分别为 a i + 1和 a i + a i + 1 。找到要对数组执行的最小操作数以使其最佳。
例子：
Input : n = 2
a[] = [1, 1]
Output : 1
Explanation:
Here, gcd(1,1) = 1. So, to make
it best we have to replace array
by [(1-1), (1+1)] = [0, 2]. Now,
gcd(0, 2) > 1. Hence, in one
operation array become best.
Input : n = 3
a[] = [6, 2, 4]
Output :0
Explanation:
Here, gcd(6,2,4) > 1.
Hence, no operation is required. 我们首先计算 gcd(array) 并检查它是否大于 1。如果是，则该数组已经是最好的，否则我们贪婪地检查是否否。通过使用以下属性使所有的移动都需要移动：当有两个奇数时，您可以在一个移动中使它们成为偶数，否则如果有一个奇数和一个偶数，那么您需要两个移动。
下面是上述方法的实现：
C++
// CPP program to find bestArray
#include
using namespace std;
// Calculating gcd of two numbers
int gcd(int a, int b){
if (a == 0)
return b;
return gcd(b%a, a);
}
void bestArray(int arr[], int n){
bool even[n] = {false};
int ans = 0;
// calculating gcd and
// counting the even numbers
for(int i = 0; i < n; i++){
ans = gcd(ans, arr[i]);
if(arr[i] % 2 == 0)
even[i] = true;
}
// check array is already best
if(ans > 1)
cout << 0 << endl;
else{
// counting the number
// of operations required.
ans = 0;
for(int i = 0; i < n-1; i++){
if(!even[i]){
even[i] = true;
even[i+1] = true;
if(arr[i+1]%2 != 0){
ans+=1;
}
else
ans+=2;
}
}
if(!even[n-1]){
ans+=2;
}
cout << ans << endl;
}
}
// driver code
int main(){
int arr[] = {57, 30, 28, 81, 88, 32, 3, 42, 25};
int n = 9;
bestArray(arr, n);
int arr1[] = {1, 1};
n = 2;
bestArray(arr1, n);
int arr2[] = {6, 2, 4};
n = 3;
bestArray(arr2, n);
}
/*This code is contributed by Sagar Shukla.*/ Java
// Java code to find best array
import java.util.*;
import java.lang.*;
public class GeeksforGeeks{
// function to calculate gcd of two numbers.
public static int gcd(int a, int b){
if (a == 0)
return b;
return gcd(b%a, a);
}
public static void bestArray(int arr[], int n){
boolean even[] = new boolean[n];
int ans = 0;
for(int i=0; i 1)
System.out.println(0);
else{
// counting the number of operations required.
ans = 0;
for(int i=0; i Python
# code to find the best array
from fractions import gcd
def bestArray(a,n):
even = [False]*n
ans = 0
# calculating the gcd and
# counting the even numbers
for i in xrange(n):
ans = gcd(ans, a[i])
if a[i]%2 == 0:
even[i] = True
# check if array is already best.
if ans > 1:
print (0)
else:
# calculating the no of
# operations required.
ans = 0
for i in xrange(n-1):
if not even[i]:
even[i] = True
even[i+1] = True
if a[i+1]%2 != 0:
ans += 1
else:
ans += 2
if not even[n-1]:
ans += 2
print (ans)
# driver code
n = 9
a = [57, 30, 28, 81, 88, 32, 3, 42, 25]
bestArray(a,n)
n = 2
a = [1, 1]
bestArray(a,n)
n = 3
a = [6, 2, 4]
bestArray(a,n)C#
// C# code to find best array
using System;
public class GFG {
// function to calculate gcd
// of two numbers.
public static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
public static void bestArray(int []arr, int n)
{
bool []even = new bool[n];
int ans = 0;
for(int i = 0; i < n; i++)
even[i] = false;
// calculating gcd and
// counting the even numbers
for(int i = 0; i < n; i++)
{
ans = gcd(ans, arr[i]);
if(arr[i] % 2 == 0)
even[i] = true;
}
// check array is already best
if(ans > 1)
Console.WriteLine(0);
else
{
// counting the number of
// operations required.
ans = 0;
for(int i = 0; i < n-1; i++)
{
if(!even[i])
{
even[i] = true;
even[i+1] = true;
if(arr[i+1] % 2 != 0)
ans += 1;
else
ans += 2;
}
}
if(!even[n-1])
ans += 2;
Console.WriteLine(ans);
}
}
// driver code
public static void Main()
{
int []arr = {57, 30, 28, 81, 88,
32, 3, 42, 25};
int n = 9;
bestArray(arr, n);
int []arr1 = {1, 1};
n = 2;
bestArray(arr1, n);
int []arr2 = {6, 2, 4};
n = 3;
bestArray(arr2, n);
}
}
// This code is contributed by vt_m.Javascript
输出：
8
1
0