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
Solution 1
| 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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23 | 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;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23 | 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;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19 | 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
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19 | 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;
}
|