题解 | #快速幂#
快速幂
https://www.nowcoder.com/practice/defdedf4fe984c6c91eefa6b00d5f4f0
def ksm(a, b, p): # mod p 下的快速幂
res = 1
while b != 0:
if b & 1 == 1:
res = (res * a) % p
a = (a * a) % p
b >>= 1
return res
q = int(input())
for _ in range(q):
a, b, p = map(int, input().split())
print(ksm(a, b, p))
