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372. Super Pow

Description

Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.

 

Example 1:

Input: a = 2, b = [3]
Output: 8

Example 2:

Input: a = 2, b = [1,0]
Output: 1024

Example 3:

Input: a = 1, b = [4,3,3,8,5,2]
Output: 1

 

Constraints:

• 1 <= a <= 231 - 1
• 1 <= b.length <= 2000
• 0 <= b[i] <= 9
• b does not contain leading zeros.

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      private final int mod = 1337;
  
      public int superPow(int a, int[] b) {
          long ans = 1;
          for (int i = b.length - 1; i >= 0; --i) {
              ans = ans * qpow(a, b[i]) % mod;
              a = qpow(a, 10);
          }
          return (int) ans;
      }
  
      private long qpow(long a, int n) {
          long ans = 1;
          for (; n > 0; n >>= 1) {
              if ((n & 1) == 1) {
                  ans = ans * a % mod;
              }
              a = a * a % mod;
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int superPow(int a, vector<int>& b) {
          using ll = long long;
          const int mod = 1337;
          ll ans = 1;
          auto qpow = [&](ll a, int n) {
              ll ans = 1;
              for (; n; n >>= 1) {
                  if (n & 1) {
                      ans = ans * a % mod;
                  }
                  a = a * a % mod;
              }
              return (int) ans;
          };
          for (int i = b.size() - 1; ~i; --i) {
              ans = ans * qpow(a, b[i]) % mod;
              a = qpow(a, 10);
          }
          return ans;
      }
  };
  

• class Solution:
      def superPow(self, a: int, b: List[int]) -> int:
          mod = 1337
          ans = 1
          for e in b[::-1]:
              ans = ans * pow(a, e, mod) % mod
              a = pow(a, 10, mod)
          return ans
  
  

• func superPow(a int, b []int) int {
  	const mod int = 1337
  	ans := 1
  	qpow := func(a, n int) int {
  		ans := 1
  		for ; n > 0; n >>= 1 {
  			if n&1 == 1 {
  				ans = ans * a % mod
  			}
  			a = a * a % mod
  		}
  		return ans
  	}
  	for i := len(b) - 1; i >= 0; i-- {
  		ans = ans * qpow(a, b[i]) % mod
  		a = qpow(a, 10)
  	}
  	return ans
  }
  

• function superPow(a: number, b: number[]): number {
      let ans = 1;
      const mod = 1337;
      const qpow = (a: number, n: number): number => {
          let ans = 1;
          for (; n; n >>= 1) {
              if (n & 1) {
                  ans = Number((BigInt(ans) * BigInt(a)) % BigInt(mod));
              }
              a = Number((BigInt(a) * BigInt(a)) % BigInt(mod));
          }
          return ans;
      };
      for (let i = b.length - 1; ~i; --i) {
          ans = Number((BigInt(ans) * BigInt(qpow(a, b[i]))) % BigInt(mod));
          a = qpow(a, 10);
      }
      return ans;
  }
  
  

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All Solutions