O(1)快速乘

// by zzq
inline LL Mul(LL a,LL b,LL mod){
	LL tmp=a*b-(LL)((long double)a/mod*b+0.5)*mod; tmp%=mod;
	return tmp<0?tmp+mod:tmp;
}
原文地址:https://www.cnblogs.com/ac-evil/p/11615527.html