求 K 次交替减 X 加 Y 等于 0 形成的数
原文:https://www . geesforgeks . org/find-number-by-k-times-alternative-reducing-x-and-add-y-to-0/
给定三个正整数 K 、 X 、 Y ，任务是求 X 交替减去 Y 到 0 合计 K 次数形成的数。
示例:
输入: X = 2，Y = 5，K = 3 输出: 1 说明: 以下是在 0 上执行 K(= 3)次的操作: 操作 1: 将值 0 减少 X(= 2)将其修改为 0–2 = 2。 操作 2: 将值-2 增加 Y(= 5)将其修改为-2 + 5 = 3。 操作 3: 将值 3 减少 X(= 2)将其修改为 3–2 = 1。 修改值后得到的值为 1。
输入: X = 1，Y = 100，K = 4 T3】输出: 198
天真法:给定的问题可以通过执行给定的操作 K 次数并打印得到的结果来解决。
时间复杂度: O(K) 辅助空间: O(1)
高效方法:上述方法也可以通过查找 K 移动次数中递减(使用 X 的值)和递增(使用 Y 的值)的总值，然后打印这些值的总和作为结果来优化。
必须添加到结果中的值计算如下:
addY = Y*(K/2) 其中，K/2 执行加法运算的次数。
必须从结果中减去的值计算如下:
addY = Y*(K/2 + K&1) 其中，K/2 执行减法运算的次数如果运算次数为奇数，则执行额外的减法运算。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the value obtained
// after alternatively reducing X and
// adding Y to 0 total K number of times
int positionAfterKJumps(int X, int Y, int K)
{
// Stores the final result after
// adding only Y to 0
int addY = Y * (K / 2);
// Stores the final number after
// reducing only X from 0
int reduceX = -1 * X * (K / 2 + K % 2);
// Return the result obtained
return addY + reduceX;
}
// Driver Code
int main()
{
int X = 2, Y = 5, K = 3;
cout << positionAfterKJumps(X, Y, K);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG {
// Function to find the value obtained
// after alternatively reducing X and
// adding Y to 0 total K number of times
static int positionAfterKJumps(int X, int Y, int K)
{
// Stores the final result after
// adding only Y to 0
int addY = Y * (K / 2);
// Stores the final number after
// reducing only X from 0
int reduceX = -1 * X * (K / 2 + K % 2);
// Return the result obtained
return addY + reduceX;
}
// Driver Code
public static void main(String[] args) {
int X = 2, Y = 5, K = 3;
System.out.print(positionAfterKJumps(X, Y, K));
}
}
// This code is contributed by code_hunt.
Python 3
# Python program for the above approach
# Function to find the value obtained
# after alternatively reducing X and
# adding Y to 0 total K number of times
def positionAfterKJumps(X, Y, K):
# Stores the final result after
# adding only Y to 0
addY = Y * (K // 2)
# Stores the final number after
# reducing only X from 0
reduceX = -1 * X * (K // 2 + K % 2)
# Return the result obtained
return addY + reduceX
# Driver Code
X = 2
Y = 5
K = 3
print(positionAfterKJumps(X, Y, K))
# This code is contributed by subhammahato348.
C
// C# program for the above approach
using System;
class GFG {
// Function to find the value obtained
// after alternatively reducing X and
// adding Y to 0 total K number of times
static int positionAfterKJumps(int X, int Y, int K)
{
// Stores the final result after
// adding only Y to 0
int addY = Y * (K / 2);
// Stores the final number after
// reducing only X from 0
int reduceX = -1 * X * (K / 2 + K % 2);
// Return the result obtained
return addY + reduceX;
}
// Driver Code
public static void Main(String[] args)
{
int X = 2, Y = 5, K = 3;
Console.WriteLine(positionAfterKJumps(X, Y, K));
}
}
// This code is contributed by ukasp.
java 描述语言
<script>
// JavaScript Program to implement
// the above approach
// Function to find the value obtained
// after alternatively reducing X and
// adding Y to 0 total K number of times
function positionAfterKJumps(X, Y, K) {
// Stores the final result after
// adding only Y to 0
let addY = Y * Math.floor(K / 2);
// Stores the final number after
// reducing only X from 0
let reduceX = -1 * X * (Math.floor(K / 2) + K % 2);
// Return the result obtained
return addY + reduceX;
}
// Driver Code
let X = 2, Y = 5, K = 3;
document.write(positionAfterKJumps(X, Y, K));
// This code is contributed by Potta Lokesh
</script>
Output:
1
时间复杂度:O(1) T5辅助空间:** O(1)
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