给定一个简单的带有两个锅的加权秤，我们得到一个权重T和其他一些权重，这些权重是特定数字a的幂，我们的目标是使用给定的权重来平衡这些锅。更正式地说，我们需要满足这个方程，
T +(a的一些幂)=(a的一些其他幂)
请记住，我们恰好被赋予与一种能力相对应的权重。
例子：
T = 11 and a = 4
Then scale can be balanced as,
11 + 4 + 1 = 16
我们可以看到，在这个问题中，我们的第一个目标是用a的幂表示T，所以当我们用a的底数写T时，如果表示只有1和0，那么我们知道对应于1的权重就可以使T 。 例如，
If T is 10 and a is 3 then representing 10 on
base of 3 we get 101 i.e. using 0th power of 3
and 2nd power of 3 (1 + 9) we can get 10
现在，如果基本表示形式的所有数字都不是1或0，那么我们就无法平衡比例，除非在基本表示形式中某个数字是(a – 1)的情况下，因为在这种情况下，我们可以将相应的幂转移到T的那一侧并将基本表示形式中的左侧数字增加1。例如，
If T is 7 and a is 3 then representing 7 on base
3 we get 021.
Now while looping over representation from right
to left we get 2 (which is 3 - 1), in this case,
we can transfer corresponding power(3^1) to T’s
side and increase the left number by 1 i.e.,
T’s side 7 + 3 = 10 and representation 101 (9 + 1)
which has only 1s and 0s now, so scale can be balanced.
因此，现在解决此问题的算法可以表示如下，以a的底数表示T，如果基本表示的所有数字均为1或0，则可以平衡小数位数，否则，如果任何数字为(a，则从表示的右侧循环– 1)，然后将左侧数字增加1，继续执行此操作，并忽略1、0，(a – 1)和一个表示形式的个案。如果处理了完整的基本表示，则可以平衡比例，否则无法平衡。
C++
// C++ code to check whether scale can be balanced or not
#include
using namespace std;
// method returns true if balancing of scale is possible
bool isBalancePossible(int T, int a)
{
// baseForm vector will store T's representation on
// base a in reverse order
vector baseForm;
// convert T to representation on base a
while (T) {
baseForm.push_back(T % a);
T /= a;
}
// make first digit of representation as 0
baseForm.push_back(0);
// loop over base representation of T
for (int i = 0; i < baseForm.size(); i++) {
// if any digit is not 0, 1, (a - 1) or a
// then balancing is not possible
if (baseForm[i] != 0 && baseForm[i] != 1 &&
baseForm[i] != (a - 1) && baseForm[i] != a)
return false;
// if digit is a or (a - 1) then increase left
// index's count/ (case, when this weight is
// transferred to T's side)
if (baseForm[i] == a || baseForm[i] == (a - 1))
baseForm[i + 1] += 1;
}
// if representation is processed then balancing
// is possible
return true;
}
// Driver code to test above methods
int main()
{
int T = 11;
int a = 4;
bool balancePossible = isBalancePossible(T, a);
if (balancePossible)
cout << "Balance is possible" << endl;
else
cout << "Balance is not possible" << endl;
return 0;
} Java
// Java code to check whether
// scale can be balanced or not
import java.util.*;
class GFG
{
// method returns true if balancing
// of scale is possible
static boolean isBalancePossible(int T, int a)
{
// baseForm vector will store T's
// representation on base a
// in reverse order
Vector baseForm = new Vector<>();
int s = 0;
// convert T to representation on base a
while (T > 0)
{
baseForm.add(T % a);
T /= a;
s++;
}
// make first digit of representation as 0
baseForm.add(0);
// loop over base representation of T
for (int i = 0; i < s; i++)
{
// if any digit is not 0, 1, (a - 1) or a
// then balancing is not possible
if (baseForm.get(i) != 0 &&
baseForm.get(i) != 1 &&
baseForm.get(i) != (a - 1) &&
baseForm.get(i) != a)
{
return false;
}
// if digit is a or (a - 1) then increase left
// index's count/ (case, when this weight is
// transferred to T's side)
if (baseForm.get(i) == a || baseForm.get(i) == (a - 1))
{
baseForm.add(i + 1, baseForm.get(i + 1) + 1);
}
}
// if representation is processed
// then balancing is possible
return true;
}
// Driver code
public static void main(String[] args)
{
int T = 11;
int a = 4;
boolean balancePossible = isBalancePossible(T, a);
if (balancePossible)
{
System.out.println("Balance is possible");
}
else
{
System.out.println("Balance is not possible");
}
}
}
// This code has been contributed by 29AjayKumar C#
// C# code to check whether
// scale can be balanced or not
using System;
using System.Collections.Generic;
class GFG
{
// method returns true if balancing
// of scale is possible
static bool isBalancePossible(int T, int a)
{
// baseForm vector will store T's
// representation on base a
// in reverse order
List baseForm = new List();
int s = 0;
// convert T to representation on base a
while (T > 0)
{
baseForm.Add(T % a);
T /= a;
s++;
}
// make first digit of representation as 0
baseForm.Add(0);
// loop over base representation of T
for (int i = 0; i < s; i++)
{
// if any digit is not 0, 1, (a - 1) or a
// then balancing is not possible
if (baseForm[i] != 0 &&
baseForm[i] != 1 &&
baseForm[i] != (a - 1) &&
baseForm[i] != a)
{
return false;
}
// if digit is a or (a - 1) then increase left
// index's count/ (case, when this weight is
// transferred to T's side)
if (baseForm[i] == a || baseForm[i] == (a - 1))
{
baseForm.Insert(i + 1, baseForm[i+1] + 1);
}
}
// if representation is processed
// then balancing is possible
return true;
}
// Driver code
public static void Main()
{
int T = 11;
int a = 4;
bool balancePossible = isBalancePossible(T, a);
if (balancePossible)
{
Console.WriteLine("Balance is possible");
}
else
{
Console.WriteLine("Balance is not possible");
}
}
}
/* This code contributed by PrinciRaj1992 */ PHP
输出：
Balance is possible