给定五个整数X , Y , A , B和N ,任务是通过精确地执行N次以下操作来找到X和Y之间的最大可能绝对差:
- 将X的值减1,直到A。
- 将Y的值减1,直到B。
注意: (X – A + Y – B)的值必须大于或等于N
例子:
Input: X = 12, Y = 8, A = 8, B = 7, N = 2
Output: 4
Explanation:
Decrementing the value of X by 1. Therefore, X = X – 1 = 11
Decrementing the value of Y by 1. Therefore, Y = Y – 1 = 7
Therefore, the maximum absolute difference between X and Y = abs(X – Y) = abs(11 – 7) = 4
Input: X = 10, Y = 10, A = 8, B = 5, N = 3
Output: 3
Explanation:
Decrementing the value of Y by 1 three times. Therefore, Y = Y – 3 = 7
Therefore, the maximum absolute difference between X and Y = abs(X – Y) = abs(10 – 7) = 3
方法:可以使用贪婪技术解决问题。请按照以下步骤解决问题:
- 初始化一个变量,例如n1,以存储在X上执行的最大操作数。
- 更新n1 = min(N,X – A) 。
- 初始化一个变量,例如n2,以存储在Y上执行的最大操作数。
- 更新n2 = min(N,Y – B) 。
- 初始化变量diff_X_Y_1来存储X和Y的绝对差,方法是先将X的值精确地减少1 (最小为N(n,n1)次) ,然后再将Y的值减少余下的操作次数。
- 初始化变量diff_X_Y_2来存储X和Y的绝对差,方法是先将Y的值精确地减少1 (最小N(n,n2)次) ,然后再将X的值减少余下的操作次数。
- 最后,输出max(diff_X_Y_1,diff_X_Y_2)的值。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the absolute difference
// between X and Y with given operations
int AbsDiff(int X, int Y, int A, int B, int N)
{
// Stores maximum absolute difference
// between X and Y with given operations
int maxDiff = 0;
// Stores maximum number of operations
// performed on X
int n1 = X - A;
// Update X
X = X - min(N, n1);
// Decrementing N at most N times
N = N - min(N, n1);
// Stores maximum number of operations
// performed on Y
int n2 = Y - B;
// Update Y
Y = Y - min(N, n2);
// Update maxDiff
maxDiff = abs(X - Y);
return maxDiff;
}
// Function to find the max absolute difference
// between X and Y with given operations
int maxAbsDiff(int X, int Y, int A, int B, int N)
{
// Stores absolute difference between
// X and Y by first decrementing X and then Y
int diffX_Y_1;
// Stores absolute difference between X
// and Y first decrementing Y and then X
int diffX_Y_2;
// Update diffX_Y_1
diffX_Y_1 = AbsDiff(X, Y, A, B, N);
// Swap X, Y and A, B
swap(X, Y);
swap(A, B);
// Update diffX_Y_2
diffX_Y_2 = AbsDiff(X, Y, A, B, N);
return max(diffX_Y_1, diffX_Y_2);
}
// Driver Code
int main()
{
int X = 10;
int Y = 10;
int A = 8;
int B = 5;
int N = 3;
cout << maxAbsDiff(X, Y, A, B, N);
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to find the absolute difference
// between X and Y with given operations
public static int AbsDiff(int X, int Y, int A,
int B, int N)
{
// Stores maximum absolute difference
// between X and Y with given operations
int maxDiff = 0;
// Stores maximum number of operations
// performed on X
int n1 = X - A;
// Update X
X = X - Math.min(N, n1);
// Decrementing N at most N times
N = N - Math.min(N, n1);
// Stores maximum number of operations
// performed on Y
int n2 = Y - B;
// Update Y
Y = Y - Math.min(N, n2);
// Update maxDiff
maxDiff = Math.abs(X - Y);
return maxDiff;
}
// Function to find the max absolute difference
// between X and Y with given operations
public static int maxAbsDiff(int X, int Y, int A,
int B, int N)
{
// Stores absolute difference between
// X and Y by first decrementing X and then Y
int diffX_Y_1;
// Stores absolute difference between X
// and Y first decrementing Y and then X
int diffX_Y_2;
// Update diffX_Y_1
diffX_Y_1 = AbsDiff(X, Y, A, B, N);
// Swap X, Y and A, B
int temp1 = X;
X = Y;
Y = temp1;
int temp2 = A;
A = B;
B = temp2;
// Update diffX_Y_2
diffX_Y_2 = AbsDiff(X, Y, A, B, N);
return Math.max(diffX_Y_1, diffX_Y_2);
}
// Driver code
public static void main(String[] args)
{
int X = 10;
int Y = 10;
int A = 8;
int B = 5;
int N = 3;
System.out.println(maxAbsDiff(X, Y, A, B, N));
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program to implement
# the above approach
# Function to find the absolute difference
# between X and Y with given operations
def AbsDiff(X, Y, A, B, N):
# Stores maximum absolute difference
# between X and Y with given operations
maxDiff = 0
# Stores maximum number of operations
# performed on X
n1 = X - A
# Update X
X = X - min(N, n1)
# Decrementing N at most N times
N = N - min(N, n1)
# Stores maximum number of operations
# performed on Y
n2 = Y - B
# Update Y
Y = Y - min(N, n2)
# Update maxDiff
maxDiff = abs(X - Y)
return maxDiff
# Function to find the max absolute difference
# between X and Y with given operations
def maxAbsDiff(X, Y, A, B, N):
# Stores absolute difference between
# X and Y by first decrementing X and then Y
diffX_Y_1 = AbsDiff(X, Y, A, B, N)
# Swap X, Y and A, B
temp1 = X
X = Y
Y = temp1
temp2 = A
A = B
B = temp2
# Stores absolute difference between X
# and Y first decrementing Y and then X
diffX_Y_2 = AbsDiff(X, Y, A, B, N)
return max(diffX_Y_1, diffX_Y_2)
# Driver Code
X = 10
Y = 10
A = 8
B = 5
N = 3
print(maxAbsDiff(X, Y, A, B, N))
# This code is contributed by sanjoy_62
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to find the absolute difference
// between X and Y with given operations
public static int AbsDiff(int X, int Y, int A,
int B, int N)
{
// Stores maximum absolute difference
// between X and Y with given operations
int maxDiff = 0;
// Stores maximum number of operations
// performed on X
int n1 = X - A;
// Update X
X = X - Math.Min(N, n1);
// Decrementing N at most N times
N = N - Math.Min(N, n1);
// Stores maximum number of operations
// performed on Y
int n2 = Y - B;
// Update Y
Y = Y - Math.Min(N, n2);
// Update maxDiff
maxDiff = Math.Abs(X - Y);
return maxDiff;
}
// Function to find the max absolute difference
// between X and Y with given operations
public static int maxAbsDiff(int X, int Y, int A,
int B, int N)
{
// Stores absolute difference between
// X and Y by first decrementing X and then Y
int diffX_Y_1;
// Stores absolute difference between X
// and Y first decrementing Y and then X
int diffX_Y_2;
// Update diffX_Y_1
diffX_Y_1 = AbsDiff(X, Y, A, B, N);
// Swap X, Y and A, B
int temp1 = X;
X = Y;
Y = temp1;
int temp2 = A;
A = B;
B = temp2;
// Update diffX_Y_2
diffX_Y_2 = AbsDiff(X, Y, A, B, N);
return Math.Max(diffX_Y_1, diffX_Y_2);
}
// Driver code
public static void Main(String[] args)
{
int X = 10;
int Y = 10;
int A = 8;
int B = 5;
int N = 3;
Console.WriteLine(maxAbsDiff(X, Y, A, B, N));
}
}
// This code is contributed by Amit Katiyar
Javascript
3
时间复杂度: O(1)
辅助空间: O(1)