检查是否可以使用秤和一些砝码测量物品

给定一些质量权重a ⁰ ，a ¹ ，a ² ，…， ¹⁰⁰ ，a为整数和一个可以在秤的两边放置砝码的秤。检查是否可以使用这些重量和秤来测量重量W的特定项目。

Constraints: 2 ≤ W ≤ 10⁹

例子：

Input : a = 2, W = 5
Output : YES
Explanation : Weights of masses (2⁰ = 1) and (2² = 4) can be placed on one side of the scale and the item can be placed on other side, i.e. 1 + 4 = 5

Input : a = 4, W = 11
Output : YES
Explanation : Weights of masses (4⁰ = 1) and (4¹ = 4) and the item can be placed on one side and a weight of mass (4² = 16) can be placed on the other side, i.e. 1 + 4 + 11 = 16

Input : a = 4, W = 7
Output : NO

方法：

首先，可以仔细观察到，对于a = 2或a = 3，答案始终存在。

保持包含质量权重的数组，并且仅包括小于10 ^9的权重

现在，可以通过将当前权重包括到包含
物品或包括当前重量到包含物品的另一侧或根本不使用该重量。

下面是上述方法的实现。

C++

// CPP Program to check whether an item
// can be measured using some weight and
// a weighing scale.
#include 
using namespace std;
  
// Variable to denote that answer has
// been found
int found = 0;
  
void solve(int idx, int itemWt, int wts[],
           int N)
{
    if (found)
        return;
  
    // Item has been measured
    if (itemWt == 0) {
        found = 1;
        return;
    }
  
    // If the index of current weight
    // is greater than totalWts
    if (idx > N)
        return;
  
    // Current weight is not included
    // on either side
    solve(idx + 1, itemWt, wts, N);
  
    // Current weight is included on the
    // side containing item
    solve(idx + 1, itemWt + wts[idx], wts,
          N);
  
    // Current weight is included on the
    // side opposite to the side
    // containing item
    solve(idx + 1, itemWt - wts[idx], wts,
          N);
}
  
// This function computes the required array
// of weights using a
bool checkItem(int a, int W)
{
    // If the a is 2 or 3, answer always
    // exists
    if (a == 2 || a == 3)
        return 1;
  
    int wts[100]; // weights array
    int totalWts = 0; // feasible weights
    wts[0] = 1;
    for (int i = 1;; i++) {
        wts[i] = wts[i - 1] * a;
        totalWts++;
  
        // if the current weight
        // becomes greater than 1e9
        // break from the loop
        if (wts[i] > 1e9)
            break;
    }
    solve(0, W, wts, totalWts);
    if (found)
        return 1;
  
    // Item can't be measured
    return 0;
}
  
// Driver Code to test above functions
int main()
{
    int a = 2, W = 5;
    if (checkItem(a, W))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
  
    a = 4, W = 11, found = 0;
    if (checkItem(a, W))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
  
    a = 4, W = 7, found = 0;
    if (checkItem(a, W))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

Java

// Java Program to check whether an item
// can be measured using some weight and
// a weighing scale.
  
class GFG {
  
// Variable to denote that answer has
// been found
    static int found = 0;
  
    static void solve(int idx, int itemWt, int wts[], int N) {
        if (found == 1) {
            return;
        }
  
        // Item has been measured
        if (itemWt == 0) {
            found = 1;
            return;
        }
  
        // If the index of current weight
        // is greater than totalWts
        if (idx > N) {
            return;
        }
  
        // Current weight is not included
        // on either side
        solve(idx + 1, itemWt, wts, N);
  
        // Current weight is included on the
        // side containing item
        solve(idx + 1, itemWt + wts[idx], wts,
                N);
  
        // Current weight is included on the
        // side opposite to the side
        // containing item
        solve(idx + 1, itemWt - wts[idx], wts,
                N);
    }
  
// This function computes the required array
// of weights using a
    static boolean checkItem(int a, int W) {
        // If the a is 2 or 3, answer always
        // exists
        if (a == 2 || a == 3) {
            return true;
        }
  
        int wts[] = new int[100]; // weights array
        int totalWts = 0; // feasible weights
        wts[0] = 1;
        for (int i = 1;; i++) {
            wts[i] = wts[i - 1] * a;
            totalWts++;
  
            // if the current weight
            // becomes greater than 1e9
            // break from the loop
            if (wts[i] > 1e9) {
                break;
            }
        }
        solve(0, W, wts, totalWts);
        if (found == 1) {
            return true;
        }
  
        // Item can't be measured
        return false;
    }
  
// Driver Code to test above functions
    public static void main(String[] args) {
  
        int a = 2, W = 5;
        if (checkItem(a, W)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
  
        a = 4; W = 11;found = 0;
        if (checkItem(a, W)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
  
        a = 4; W = 7; found = 0;
        if (checkItem(a, W)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}
  
//this code contributed by Rajput-Ji

C#

// C# Program to check whether an item
// can be measured using some weight and
// a weighing scale.
using System; 
public class GFG {
   
// Variable to denote that answer has
// been found
    static int found = 0;
   
    static void solve(int idx, int itemWt, int []wts, int N) {
        if (found == 1) {
            return;
        }
   
        // Item has been measured
        if (itemWt == 0) {
            found = 1;
            return;
        }
   
        // If the index of current weight
        // is greater than totalWts
        if (idx > N) {
            return;
        }
   
        // Current weight is not included
        // on either side
        solve(idx + 1, itemWt, wts, N);
   
        // Current weight is included on the
        // side containing item
        solve(idx + 1, itemWt + wts[idx], wts,
                N);
   
        // Current weight is included on the
        // side opposite to the side
        // containing item
        solve(idx + 1, itemWt - wts[idx], wts,
                N);
    }
   
// This function computes the required array
// of weights using a
    static bool checkItem(int a, int W) {
        // If the a is 2 or 3, answer always
        // exists
        if (a == 2 || a == 3) {
            return true;
        }
   
        int []wts = new int[100]; // weights array
        int totalWts = 0; // feasible weights
        wts[0] = 1;
        for (int i = 1;; i++) {
            wts[i] = wts[i - 1] * a;
            totalWts++;
   
            // if the current weight
            // becomes greater than 1e9
            // break from the loop
            if (wts[i] > 1e9) {
                break;
            }
        }
        solve(0, W, wts, totalWts);
        if (found == 1) {
            return true;
        }
   
        // Item can't be measured
        return false;
    }
   
// Driver Code to test above functions
    public static void Main() {
   
        int a = 2, W = 5;
        if (checkItem(a, W)) {
            Console.WriteLine("YES");
        } else {
            Console.WriteLine("NO");
        }
   
        a = 4; W = 11;found = 0;
        if (checkItem(a, W)) {
            Console.WriteLine("YES");
        } else {
            Console.WriteLine("NO");
        }
   
        a = 4; W = 7; found = 0;
        if (checkItem(a, W)) {
            Console.WriteLine("YES");
        } else {
            Console.WriteLine("NO");
        }
    }
}
   
//this code contributed by Rajput-Ji

输出：

YES
YES
NO

时间复杂度： O(3 ^N )，其中N不能大于20，因为4 ²⁰大于10 ⁹

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