解：由题意可得$$a+b=-20$$，$$ab=1$$，$$c+d=17$$，$$cd=1$$  $$\therefore \left( a+c \right)\left( b+c \right)\left( a-d \right)\left( b-d \right)$$  $$=\left[ ab+\left( a+b \right)c+{{c}^{2}} \right]\left[ ab-\left( a+b \right)d+{{d}^{2}} \right]$$  $$=\left( 1-20c+{{c}^{2}} \right)\left( 1+20d+{{d}^{2}} \right)$$  $$=1+20d+{{d}^{2}}-20c-400-20d+{{c}^{2}}+20c+1$$  $$={{d}^{2}}+{{c}^{2}}+2-400$$  $$={{\left( c+d \right)}^{2}}-400$$  $$={{17}^{2}}-400$$  $$=-111$$，  故选：$$\text{D}$$． 
设$$a$$，$$b$$是方程$${{x}^{2}}+20x+1=0$$的两个根，$$c$$，$$d$$是方程$${{x}^{2}}-17x+1=0$$的两个根，则代数式$$\left( a+c \right)\left( b+c \right)\left( a-d \right)\left( b-d \right)$$的值为．

A. $$-2017$$ 
B. $$0$$ 
C. $$340$$ 
D. $$-111$$ 