题解 | #快速幂#
快速幂
https://www.nowcoder.com/practice/defdedf4fe984c6c91eefa6b00d5f4f0
# https://zhuanlan.zhihu.com/p/95902286 """ 重要的是2点: #####s1: a^n = (1) n是奇数, a^(n-1) x a (2) 偶数, (a^n/2) ^2 ######s2: 取模的公式: (aXb)mod c = (a mod c x b mod c)mod c """ times = int(input()) listall = [] def quickpow(a,b,p): if b == 0: return 1 if b%2==1: return quickpow(a, b-1,p) *a % p else: # 此步骤tmp不可省略,因为计算量减半 tmp = quickpow(a,b/2,p)%p return tmp*tmp%p for i in range(times): a,b,p = map(int, input().split()) elema = quickpow(a,b,p) res = elema%p print(res)