HDOJ-3343 An ant's story(積分)
橡皮筋均勻伸長m m/s,螞蟻勻速從一端以v m/s走向另一端,問螞蟻何時走到終點.
變換位移的方法: 設相對位移x,有比值關係vdtdx=L+mtL,解: L=∫0Ldx=∫0TL+mtLvdt=mLvln(L+mt)∣∣∣0T=mLvlnLL+mT
所以,T=mL(em/v−1)
暴力方法:
設到終點的距離x(t),
dx=L+mtxmdt−vdt
令T=t+mL,整理得x′−T
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