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By Vieta's formulas, $$\begin{array}{rl}c+d& =10a,\\ cd& =-11b,\\ a+b& =10c,\\ ab& =-11d.\end{array}$$From the first equation, $$d=10a-c.$$From the third equation, $$b=10c-a.$$Substituting into the second and fourth equations, we get $$\begin{array}{rl}c(10a-c)& =-11(10c-a),\\ a(10c-a)& =-11(10a-c).\end{array}$$Expanding, we get $$\begin{array}{rl}10ac-c^2& =-110c+11a,\\ 10ac-a^2& =-110a+11c.\end{array}$$Subtracting these equations, we get $$a^2-c^2=121a-121c,$$so $(a+c)(a-c)=121(a-c).$ Since $a$ and $c$ are distinct, we can divide both sides by $a-c,$ to get $$a+c=121.$$Hence, $a+b+c+d=10c+10a=10(a+c)=\boxed{1210}.$