检查是否可以通过给定的操作从数组中获得和 Y

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一个具有总和X的序列，使得子序列的每个元素的总和乘以一个数组元素等于Y .
注意：这里 X 总是小于 Y。

例子：

Input: arr[] = {1, 2, 7, 9, 10}, X = 11, Y = 13 
Output: Yes 
Explanation: 
The given value of X( = 11) can be split into a sequence {9, 2} such that 9 * 1(= arr[0] + 2 * 2(= arr[1]) = 13( = Y)
Input: arr[] ={1, 3, 5, 7}, X = 27, Y = 34 
Output: No

方法：按照以下步骤解决问题：

• 计算Y和X之间的差值。
• 对于每个大于 1 的数组元素arr[i] ，更新(Y – X) % (arr[i] – 1) 。
• 如果差值减少到 0，则打印“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 Code
int 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

// Java program to implement
// the above approach
class 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 Code
public 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

Python3

# 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 Code
arr = [ 1, 2, 7, 9, 10 ]
n = len(arr)
X, Y = 11, 13
 
# Function call
solve(arr, n, X, Y)
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
 
class 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)
        Console.Write("Yes");
    else
        Console.Write("No");
}
 
// Driver Code
public 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

Javascript

     
// Javascript 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[]
function solve(arr, n, X, Y)
{
 
    // Store the difference
    var diff = Y - X;
 
    // Iterate over the array
    for(var i = 0; i < n; i++)
    {
         
        if (arr[i] != 1)
        {
            diff = diff % (arr[i] - 1);
        }
    }
 
    // If diff reduced to 0
    if (diff == 0)
        document.write("Yes");
    else
        document.write("No");
}
 
// Driver Code
var arr = [ 1, 2, 7, 9, 10 ];
var n = arr.length;
var X = 11, Y = 13;
 
solve(arr, n, X, Y);
 
// This code is contributed by rutvik_56
 

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)