long long fpow(long long a,long ,long b,long long mod) { long long res = 1; while(b) { if(b & 1) res = (res * a) % mod; a = (a * a) % mod; b >>= 1; } return res; }
算法学习(3)：快速幂
原文：https://www.cnblogs.com/xiaoxingaa/p/14722862.html
