1. 程式人生 > >HDOJ-3343 An ant's story(積分)

HDOJ-3343 An ant's story(積分)

橡皮筋均勻伸長m m/sm \text{ m/s},螞蟻勻速從一端以v m/sv \text{ 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xx,有比值關係dxvdt=LL+mt\displaystyle \frac{dx}{vdt}= \frac{L}{L+mt},解: L=0Ldx=0TLL+mtvdt=Lvmln(L+mt)0T=LvmlnL+mTLL=\int _0^L dx =\int _0 ^T \frac{L}{L+mt} vdt=\frac{Lv}{m}\ln (L+mt)\Big| _0^T=\frac{Lv}{m}\ln \frac{L+mT}{L}

所以,T=Lm(em/v1)T=\frac{L}{m}(e^{m/v}-1)

暴力方法: 設到終點的距離x(t)x(t), dx=xL+mtmdtvdt\displaystyle dx=\frac{x}{L+mt}mdt-vdtT=t+Lm\displaystyle T=t+\frac{L}{m},整理得xxT+v=0\displaystyle x'-\frac{x}{T}+v=0

x+v=0. 令y=xT,yT=xxT\displaystyle y=\frac{x}{T},得y'T=x'-\frac{x}{T},於是y(T)=dy=vTdT=vlnT+C\displaystyle y(T)=\int dy=\int \frac{-v}{T}dT=-v\ln T+C. 取t=0t=0,此時T=Lm,x(T)=vLmlnLm+CLm=L,C=m+vlnLm\displaystyle T=\frac{L}{m},有x(T)=-v\frac{L}{m}ln\frac{L}{m}+C\frac{L}{m}=L,所以C=m+v\ln\frac{L}{m}
. 令x(T)=vTlnT+T(m+vlnLm)=0,:\displaystyle x(T)=-vTlnT+T(m+v\ln\frac{L}{m})=0,得: T=em/v+ln(L/m)=Lmem/v.T=e^{m/v+\ln(L/m)}=\frac{L}{m}e^{m/v}. 於是t=TLm=Lm(em/v1)\displaystyle t=T-\frac{L}{m}=\frac{L}{m}(e^{m/v}-1)時螞蟻走到終點.