给定两个字符串表示的数字，判断其中一个是否为另一个的幂。

//C++ program to check if one number is power of other by using string multiply(string one, string two)
#include <bits/stdc++.h>
using namespace std;
bool isGreaterThanEqualTo(string s1, string s2){
   if (s1.size() > s2.size())
   return true;
   return (s1 == s2);
}
string multiply(string s1, string s2){
   int n = s1.size();
   int m = s2.size();
   vector<int> result(n + m, 0);
   for (int i = n - 1; i >= 0; i--)
   for (int j = m - 1; j >= 0; j--)
   result[i + j + 1] +=
   (s1[i] - '0') * (s2[j] - '0');
   int size = result.size();
   for (int i = size - 1; i > 0; i--){
      if (result[i] >= 10) {
         result[i - 1] += result[i] / 10;
         result[i] = result[i] % 10;
      }
   }
   int i = 0;
   while (i < size && result[i] == 0)
   i++;
   if (i == size)
   return "0";
   string temp;
   while (i < size){
      temp += (result[i] + '0');
      i++;
   }
   return temp;
}
string removeLeadingZeores(string s){
   int n = s.size();
   int i = 0;
   while (i < n && s[i] == '0')
   i++;
   if (i == n)
   return "0";
   string temp;
   while (i < n)
   temp += s[i++];
   return temp;
}
bool isPower(string s1, string s2){
   s1 = removeLeadingZeores(s1);
   s2 = removeLeadingZeores(s2);
   if (s1 == "0" || s2 == "0")
   return false;
   if (s1 == "1" && s2 == "1")
   return true;
   if (s1 == "1" || s2 == "1")
   return true;
   if (s1.size() > s2.size())
   return isPower(s2, s1);
   string temp = s1;
   while (!isGreaterThanEqualTo(s1, s2))
   s1 = multiply(s1, temp);
   return s1 == s2;
}
int main(){
   string s1 = "1610", s2 = "52627712618930723";
   cout << (isPower(s1, s2) ? "YES\n" : "NO\n");
   s1 = "19972001", s2 = "2022";
   cout << (isPower(s1, s2) ? "YES\n" : "NO\n");
   return 0;
}

//C++ program to check if one number is power of other by using the isPower() method
#include <bits/stdc++.h>
using namespace std;
bool isPower(int x, long int y){
   if (x == 1)
   return (y == 1);
   long int pow = 1;
   while (pow < y)
   pow *= x;
   return (pow == y);
}
int main(){
   cout << isPower(16, 07) << endl;
   cout << isPower(10, 1) << endl;
   cout << isPower(97, 10) << endl;
   cout << isPower(2, 30) << endl;
   return 0;
}

//C++ program to check if one number is power of other by using the logarithm function to calculate value and use the compare method also
#include <iostream>
#include <math.h>
using namespace std;
bool isPower(int x, int y){
   double res1 = log(y) / log(x);
   double res2 = log(y) / log(x);
   return (res1 == res2);
}
int main(){
   cout << isPower(16, 07) << endl;
   return 0;
}

//C++ program to check if one number is power of other by using the loop changing exponent with binary search
#include <iostream>
#include <cmath>
using namespace std;
bool isIntegerPower(int x, int y){
   int low = 0, high;
   int exp = 1;
   int val = y;
   while(1){
      val = pow((double)y, exp);
      if(val == x)
      return true;
      else if(val > x)
      break;
      low = exp;
      exp = exp * 2;
      high = exp;
   }
   int mid = (low + high)/2;
   while(low < high){
      val = pow((double)y, mid);
      if(val > x){
         high = mid-1;
      } else if(val == x){
         return true;
      } else if(val < x){
         low = mid+1;
      }
      mid = (low + high)/2;
   }
   return false;
}
int main(){
   cout<<isIntegerPower(1607,2);
}

//C++ program to check if one number is power of other by using the function is checkPowerofTwo() or not
#include <iostream>
#include <cmath>
using namespace std;
int checkPowerofTwo(int n);
int main(){
   int num;
   printf("Enter the number you want to test today: ");
   scanf("%d", &num);
   if (checkPowerofTwo(num) == 1)
   printf("\n%d - this number is a power of 2\n", num);
   else
   printf("\n%d - this number is not a power of 2\n", num);
   return 0;
}
int checkPowerofTwo(int x){
   if (x == 0)
   return 0;
   while( x != 1){
      if(x % 2 != 0)
      return 0;
      x /= 2;
   }
   return 1;
}

//C++ program to check if one number is power of other by using traversing method
#include <bits/stdc++.h>
using namespace std;
int countPairs(int A[], int n, int k){
   int ans = 0;
   sort(A, A + n);
   for (int i = 0; i < n; i++){
      for (int j = i + 1; j < n; j++){
         int x = 0;
         while ((A[i] * pow(k, x)) <= A[j]){
            if ((A[i] * pow(k, x)) == A[j]){
               ans++;
               break;
            }
            x++;
         }
      }
   }
   return ans;
}
int main() {
   int A[] = {3, 8, 9, 12, 16, 10,7, 1997, 2001, 18, 4, 24, 2, 6};
   int n = sizeof(A) / sizeof(A[0]);
   int k = 3;
   cout << countPairs(A, n, k);
   return 0;
}