能否通过给定的运算从数组中获得和为Y？

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检查是否可以通过指定的运算从数组中获取和Y以及Y原文：https://www.example.com/check-if-sum-y-can-be-achieved-by-specified-operation-from-array-y原文：https://www.geeksforgeeks.org/check-if-sum-y-can-be-achieved-by-specified-operation-from-array-y

检查是否可以通过给定的运算从数组中获得和Y原文:www . geeksforgeeks . org/check-if-sum-y-可以通过给定的操作从数组中获取/

给定一个整数数组 arr[] 和两个整数 X 和 Y ，任务是检查是否有可能获得具有和 X 的序列，使得子序列的每个元素乘以一个数组元素的和等于 Y 。注意:这里 X 总是小于 y。

示例:

输入: arr[] = {1，2，7，9，10}，X = 11，Y = 13输出: Yes解释:X(= 11)的给定值可以拆分为序列{9，2}，使得 9 * 1(= arr[0]+2 * 2(= arr[1])= 13(= Y)输入: arr[] ={1，3，5，7

方法:按照以下步骤解决问题:

• 计算 Y 和 X 之间的差。
• 对于每个数组元素 arr[i] ，即 > 1 ，更新(Y–X)%(arr[I]–1)。
• 如果差值减少 0，打印“是”。否则，打印“否”。

下面是上述方法的实现:

C++

// C++ Program to implement// the above approach#include using namespace std;// Function to check if it is possible// to obtain sum Y from a sequence of// sum X from the array arr[]void solve(int arr[], int n, int X, int Y){    // Store the difference    int diff = Y - X;    // Iterate over the array    for (int i = 0; i < n; i++) {        if (arr[i] != 1) {            diff = diff % (arr[i] - 1);        }    }    // If diff reduced to 0    if (diff == 0)        cout << "Yes";    else        cout << "No";}// Driver Codeint main(){    int arr[] = { 1, 2, 7, 9, 10 };    int n = sizeof(arr) / sizeof(arr[0]);    int X = 11, Y = 13;    solve(arr, n, X, Y);    return 0;}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program to implement// the above approachclass GFG{// Function to check if it is possible// to obtain sum Y from a sequence of// sum X from the array arr[]static void solve(int arr[], int n,                  int X, int Y){    // Store the difference    int diff = Y - X;    // Iterate over the array    for(int i = 0; i < n; i++)    {        if (arr[i] != 1)        {            diff = diff % (arr[i] - 1);        }    }    // If diff reduced to 0    if (diff == 0)        System.out.print( "Yes");    else        System.out.print("No");}// Driver Codepublic static void main (String []args){    int arr[] = { 1, 2, 7, 9, 10 };    int n = arr.length;    int X = 11, Y = 13;    solve(arr, n, X, Y);}}// This code is contributed by chitranayal

Python 3

# Python3 program to implement# the above approach# Function to check if it is possible# to obtain sum Y from a sequence of# sum X from the array arr[]def solve(arr, n, X, Y):    # Store the difference    diff = Y - X    # Iterate over the array    for i in range(n):        if(arr[i] != 1):            diff = diff % (arr[i] - 1)    # If diff reduced to 0    if(diff == 0):        print("Yes")    else:        print("No")# Driver Codearr = [ 1, 2, 7, 9, 10 ]n = len(arr)X, Y = 11, 13# Function callsolve(arr, n, X, Y)# This code is contributed by Shivam Singh

C

// C# program to implement// the above approachusing System;class GFG{// Function to check if it is possible// to obtain sum Y from a sequence of// sum X from the array []arrstatic void solve(int []arr, int n,                  int X, int Y){    // Store the difference    int diff = Y - X;    // Iterate over the array    for(int i = 0; i < n; i++)    {        if (arr[i] != 1)        {            diff = diff % (arr[i] - 1);        }    }    // If diff reduced to 0    if (diff == 0)        Console.Write("Yes");    else        Console.Write("No");}// Driver Codepublic static void Main(String []args){    int []arr = { 1, 2, 7, 9, 10 };    int n = arr.Length;    int X = 11, Y = 13;    solve(arr, n, X, Y);}}// This code is contributed by Amit Katiyar

java 描述语言

Output: 

Yes

时间复杂度:O(N)T5辅助空间:** O(1)