字符串，先反转，再边删边用pos函数

var
k,a:integer;
s:string;
x:char;
function qs(n:string;k:integer;):char;
begin
qs:=n[length(n)-k+1];
end;
begin
a:=0; s:=''; read(x);
while x<>' ' do begin
s:=s+x;
read(x);
end;
readln(k);
a:=ord(qs(s,k))-48+a;
s:=''; read(x);
while x<>' ' do begin
s:=s+x;
read(x);
end;
readln(k);
a:=ord(qs(s,k))-48+a;
writeln(a);
end.

数据不大的话先整除后取余更优雅

数据不大
var
sum,b:integer;
a:longint;
function qs(n:longint;k:integer):integer;
var
j,a:integer;
begin
j:=1;
for a:=1 to k-1 do j:=j*10;
qs:=n mod (j*10);
qs:=qs div j;
end;
begin
readln(a,b);
sum:=qs(a,b);
readln(a,b);
sum:=sum+qs(a,b);
writeln(sum);
readln;
end.
数据大
var
s:string;
i,j:integer;
function qs(n:string;k:integer):integer;
begin
val(n[length(n)-k+1],qs);
end;
begin
readln(s,j);
i:=qs(s,j);
readln(s,j);
i:=i+qs(s,j);
writeln(i);
readln;
end.