算法学习(3)：快速幂

快速幂

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;
}