一列快车长 $306\text{m}$，一列慢车长 $344\text{m}$，如果两车相向而行，从相遇到离开需要 $13\text{s}$；如果两车同向而行，快车从追上慢车到超过慢车需要 $65\text{s}$，求快车、慢车各自的速度.若设快车的速度是 $x\text{m}/\text{s}$，慢车的速度是 $y\text{m}/\text{s}$，那么列出方程组为 （ ）

A．	$\mathrm{}\left\{\begin{array}{c}13\left(x+y\right)=306+344\\ 65\left(x-y\right)=344\end{array}\right.$	B．	$\mathrm{}\left\{\begin{array}{c}13\left(x+y\right)=306+344\\ 65\left(x-y\right)=344-306\end{array}\right.$
C．	$\mathrm{}\left\{\begin{array}{c}13\left(x+y\right)=306+344\\ 65\left(x-y\right)=306-344\end{array}\right.$	D．	$\mathrm{}\left\{\begin{array}{c}344y=306x\\ 65\left(x-y\right)=306+344\end{array}\right.$