// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
}// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
}// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result := 1
}// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result := 1
// If the lowest bit of n is set, the current power of x belongs in the result.
}// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result := 1
for n > 0 {
// If the lowest bit of n is set, the current power of x belongs in the result.
if n&1 == 1 {
result *= x
}
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result := 1
for n > 0 {
// If the lowest bit of n is set, the current power of x belongs in the result.
if n&1 == 1 {
result *= x
}
// Move to the next bit: square the base and shift n right by one.
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result := 1
for n > 0 {
// If the lowest bit of n is set, the current power of x belongs in the result.
if n&1 == 1 {
result *= x
}
// Move to the next bit: square the base and shift n right by one.
x *= x
n >>= 1
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result := 1
for n > 0 {
// If the lowest bit of n is set, the current power of x belongs in the result.
if n&1 == 1 {
result *= x
}
// Move to the next bit: square the base and shift n right by one.
x *= x
n >>= 1
}
// For modular exponentiation, take % mod after each *= (result and x).
}// power returns x^n for n >= 0 in O(log n) multiplications.
func power(x, n int) int {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result := 1
for n > 0 {
// If the lowest bit of n is set, the current power of x belongs in the result.
if n&1 == 1 {
result *= x
}
// Move to the next bit: square the base and shift n right by one.
x *= x
n >>= 1
}
// For modular exponentiation, take % mod after each *= (result and x).
return result
}# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
pass# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
# x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
pass# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
# x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result = 1# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
# x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result = 1
# If the lowest bit of n is set, the current power of x belongs in the result.# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
# x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result = 1
while n > 0:
# If the lowest bit of n is set, the current power of x belongs in the result.
if n & 1 == 1:
result *= x# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
# x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result = 1
while n > 0:
# If the lowest bit of n is set, the current power of x belongs in the result.
if n & 1 == 1:
result *= x
# Move to the next bit: square the base and shift n right by one.# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
# x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result = 1
while n > 0:
# If the lowest bit of n is set, the current power of x belongs in the result.
if n & 1 == 1:
result *= x
# Move to the next bit: square the base and shift n right by one.
x *= x
n >>= 1# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
# x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result = 1
while n > 0:
# If the lowest bit of n is set, the current power of x belongs in the result.
if n & 1 == 1:
result *= x
# Move to the next bit: square the base and shift n right by one.
x *= x
n >>= 1
# For modular exponentiation, take % mod after each *= (result and x).# power returns x^n for n >= 0 in O(log n) multiplications.
def power(x, n):
# x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
result = 1
while n > 0:
# If the lowest bit of n is set, the current power of x belongs in the result.
if n & 1 == 1:
result *= x
# Move to the next bit: square the base and shift n right by one.
x *= x
n >>= 1
# For modular exponentiation, take % mod after each *= (result and x).
return result// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
// If the lowest bit of n is set, the current power of x belongs in the result.
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x, n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
return result;
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
// If the lowest bit of n is set, the current power of x belongs in the result.
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
}// power returns x^n for n >= 0 in O(log n) multiplications.
function power(x: number, n: number): number {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
let result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) === 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
return result;
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
}
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
}
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
}
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
// If the lowest bit of n is set, the current power of x belongs in the result.
}
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
}
}
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
}
}
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
}
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
}
}class Solution {
// power returns x^n for n >= 0 in O(log n) multiplications.
public long power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
return result;
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
// If the lowest bit of n is set, the current power of x belongs in the result.
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
}
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
}
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
}
}public class Solution {
// Power returns x^n for n >= 0 in O(log n) multiplications.
public long Power(long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
return result;
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
// If the lowest bit of n is set, the current power of x belongs in the result.
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
}// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
return result;
}class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
}
};class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
}
};class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
}
};class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
// If the lowest bit of n is set, the current power of x belongs in the result.
}
};class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
}
}
};class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
}
}
};class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
}
};class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
}
};class Solution {
public:
// power returns x^n for n >= 0 in O(log n) multiplications.
long long power(long long x, int n) {
// x^n = product of x^(2^b) for each set bit b of n. Start the product at 1.
long long result = 1;
while (n > 0) {
// If the lowest bit of n is set, the current power of x belongs in the result.
if ((n & 1) == 1) {
result *= x;
}
// Move to the next bit: square the base and shift n right by one.
x *= x;
n >>= 1;
}
// For modular exponentiation, take % mod after each *= (result and x).
return result;
}
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