检查是否可以从给定的一组元素中获得给定的总和

给定数字数组和一个整数 x。通过添加给定数组的元素来判断是否有可能得到 x，我们可以多次选择单个元素。对于给定的数组，可以有许多求和查询。
例子：

Input : arr[] = { 2, 3}
         q[]  = {8, 7}
Output : Yes Yes
Explanation : 
2 + 2 + 2 + 2 = 8
2 + 2 + 3 = 7

这个想法是首先对给定的数组进行排序，然后使用类似于埃拉托色尼筛法的概念。首先取一个大数组（这是 x 的最大大小）。最初在它的所有索引中保持零。在零索引处设为 1（无论数组是什么，我们都可以得到零）。现在，遍历整个数组并将所有可能的值设为 1。

C++

// CPP program to find if we can get given
// sum using elements of given array.
#include 
using namespace std;
 
// maximum size of x
const int MAX = 1000;
 
// to check whether x is possible or not
int ispossible[MAX];
 
void check(int arr[], int N)
{
    ispossible[0] = 1;
    sort(arr, arr + N);
 
    for (int i = 0; i < N; ++i) {
        int val = arr[i];
 
        // if it is already possible
        if (ispossible[val])
            continue;
 
        // make 1 to all it's next elements
        for (int j = 0; j + val < MAX; ++j)
            if (ispossible[j])
                ispossible[j + val] = 1;
    }
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    check(arr, N);
    int x = 7;
    if (ispossible[x])
        cout << x << " is possible.";
    else
        cout << x << " is not possible.";
    return 0;
}

Java

// Java program to find if we can get given
// sum using elements of given array.
import java.util.*;
 
class solution
{
 
// maximum size of x
int MAX = 1000;
 
// to check whether x is possible or not
static int []ispossible = new int[1000];
static void check(int[] arr, int N)
{
     
    ispossible[0] = 1;
    Arrays.sort(arr);
 
    for (int i = 0; i < N; ++i) {
        int val = arr[i];
 
        // if it is already possible
        if (ispossible[val] == 1)
            continue;
 
        // make 1 to all it's next elements
        for (int j = 0; j + val < 1000; ++j)
            if (ispossible[j]== 1)
                ispossible[j + val] = 1;
    }
}
 
// Driver code
public static void main(String args[])
{
    int[] arr = { 2, 3 };
    int N = arr.length;
    check(arr, N);
    int x = 7;
    if (ispossible[x]== 1)
        System.out.println(x+" is possible.");
    else
        System.out.println(x+" is not possible.");
 
}
}
// This code is contributed by
// Shashank_Sharma

Python3

# Python3 program to find if we can get given
# sum using elements of the given array.
def check(arr, N):
 
    ispossible[0] = 1
    arr.sort()
 
    for i in range(0, N):
        val = arr[i]
 
        # if it is already possible
        if ispossible[val]:
            continue
 
        # make 1 to all it's next elements
        for j in range(0, MAX - val):
            if ispossible[j]:
                ispossible[j + val] = 1
 
# Driver code
if __name__ == "__main__":
 
    arr = [2, 3]
    N = len(arr)
     
    # maximum size of x
    MAX = 1000
     
    # to check whether x is possible or not
    ispossible = [0] * MAX
 
    check(arr, N)
    x = 7
     
    if ispossible[x]:
        print(x, "is possible.")
    else:
        print(x, "is not possible.")
     
# This code is contributed by
# Rituraj Jain

C#

// C# program to find if we can get given
// sum using elements of given array.
using System;
 
class GFG
{
     
// to check whether x is possible or not
static int []ispossible = new int[1000];
static void check(int[] arr, int N)
{
     
    ispossible[0] = 1;
    Array.Sort(arr);
 
    for (int i = 0; i < N; ++i)
    {
        int val = arr[i];
 
        // if it is already possible
        if (ispossible[val] == 1)
            continue;
 
        // make 1 to all it's next elements
        for (int j = 0; j + val < 1000; ++j)
            if (ispossible[j] == 1)
                ispossible[j + val] = 1;
    }
}
 
// Driver code
public static void Main()
{
    int[] arr = { 2, 3 };
    int N = arr.Length;
    check(arr, N);
    int x = 7;
    if (ispossible[x]== 1)
        Console.WriteLine(x + " is possible.");
    else
        Console.WriteLine(x + " is not possible.");
}
}
 
// This code is contributed by
// Akanksha Rai

Javascript

// Javascript program to find if we can get given
// sum using elements of given array.
 
// maximum size of x
let MAX = 1000;
 
// to check whether x is possible or not
let ispossible = new Array(MAX);
 
function check(arr, N)
{
    ispossible[0] = 1;
    arr.sort();
 
    for (let i = 0; i < N; ++i)
    {
        val = arr[i];
 
        // if it is already possible
        if (ispossible[val])
            continue;
 
        // make 1 to all it's next elements
        for (let j = 0; j + val < MAX; ++j)
            if (ispossible[j])
                ispossible[j + val] = 1;
    }
}
 
// Driver code
let arr = new Array( 2, 3 );
let N = arr.length;
check(arr, N);
let x = 7;
if (ispossible[x])
    document.write(x + " is possible.");
else
    document.write(x + " is not possible.");
 
// This code is contributed
// by _sauraahb_jaiswal

输出：

7 is possible.