[Statedef -2];调查计测部分 ...... [State 0, VarSet] type = VarSet trigger1 = !ishelper var(b) = var(a);记录上一F敌生命值 ignorehitpause = 1 [State 0, VarSet] type = VarSet trigger1 = !ishelper var(a) = enemynear,life;记录当前时刻敌生命值 ignorehitpause = 1 [State 0, VarSet] type = VarSet triggerall = !ishelper trigger1 = var(a) > var(b) trigger1 = var(b) > 0 trigger1 = !var(c) var(c) = gametime - 1;记录敌第一次回复时的gametime ignorehitpause = 1 [State 0, VarSet] type = VarSet triggerall = !ishelper trigger1 = var(a) > var(b) trigger1 = var(b) > 0 trigger1 = !!var(c) trigger1 = gametime - 1 > var(c) trigger1 = !var(d) var(d) = gametime - 1;记录敌第二次回复时的gametime ignorehitpause = 1 ...... [Statedef XXXXXX];时止解除用helper(回复抑制部分) ...... [State ] type = Pause triggerall = IsHelper triggerall = (root,var(d))%(root,var(c)) = 0;余数为0,说明n=0 trigger1 = root,var(c) > 0 && root,var(d) > 0 trigger1 = (gametime + 1)%(root,var(c)) = 0 time = 2 movetime = 2 ignorehitpause = 1 [State ] type = Superpause triggerall = IsHelper triggerall = (root,var(d))%(root,var(c)) = 0;余数为0,说明n=0 trigger1 = root,var(c) > 0 && root,var(d) > 0 trigger1 = (gametime + 1)%(root,var(c)) = 0 anim = -1 time = 2 movetime = 2 darken = 0 p2defmul = 1 unhittable = 0 ignorehitpause = 1 [State ] type = Pause triggerall = IsHelper triggerall = (root,var(d))%(root,var(c)) != 0;余数不为0,说明n!=0 trigger1 = root,var(c) > 0 && root,var(d) > 0 trigger1 = (gametime + 1)%(root,var(d) - root,var(c)) = root,var(c) time = 2 movetime = 2 ignorehitpause = 1 [State ] type = Superpause triggerall = IsHelper triggerall = (root,var(d))%(root,var(c)) != 0;余数不为0,说明n!=0 trigger1 = root,var(c) > 0 && root,var(d) > 0 trigger1 = (gametime + 1)%(root,var(d) - root,var(c)) = root,var(c) anim = -1 time = 2 movetime = 2 darken = 0 p2defmul = 1 unhittable = 0 ignorehitpause = 1 ... 解说:无时止解除的player被时止时不执行控制器但gametime依然随时间增加,故此。 设一自然常数a,则由gametime%x=n推导出a*x+n=gametime。 分别令a=0和a=1,得到一组二元一次方程: n=Gametime;x+n=GAMETIME 解方程组得x=GAMETIME-Gametime(Gametime为第一次回复时的记录,GAMETIME为第二次回复时的记录) 由于n可能为0且为0时生命值不方便记录,以上述方法计测时应当判断GAMETIME是否为Gametime的倍数。若是则按gametime%x=0式处理。 由于mugen的运行顺序是p1>>p2>>p1helper>>p2helper,因此p2的gametime要比p1的gametime慢一F,比p1的helper快一F。