var
a,n,s,i,b:integer;
begin
read(a,n);
b:=a;
s:=0;
for i:=1 to n do
begin
s:=s+b;
b:=10*b+a;
end;
write(s);
end.
题目：
求和
a+aa+aaa+aa..aa=?
[n个a]

var
a:char;
b,c:integer;
begin
read(a);
if a:='s' then b:=1;
if a:='r' then b:=0;
if a:='p' then b:=2;
randomize;
c:=random(3);
if c=b then write('pingju')
else if (c-b=1) or (c-b=-2) then write('you lose')
else write('you win');
end.
【石头剪刀布】

var
i:char;
num1,num2:real;
begin
read(num1);
read(i,i);
read(num2);
if i='+' then num2:=num1+num2
else if i='-' then num2:=num1-num2
else if i='*' then num2:=num1*num2
else if i='/' then num2:=num1/num2;
while i<>'=' do
begin
read(i,i);
if i='=' then break;
read(num1);
if i='+' then num2:=num1+num2
else if i='-' then num2:=num1-num2
else if i='*' then num2:=num1*num2
else if i='/' then num2:=num1/num2;
end;
write(num2:8:1);
end.
复杂的计算器问题，包括了多个运算符和多个运算量，为了简单起见，规定它按照输入的先后次序计算，而不考虑运算符的优先规则。如计算20+5*6+4+3/5=？，运算顺序为+*++/，考虑终止循环的条件是什么？与手工操作计算器

var
a:array['a'..'z'] of integer;
i:char;
begin
for i:='a' to 'z' do
a[i]:=0;
read(i);
while i<>'.' do
begin
if (i>='a') and (i<='z') then
a[i]:=a[i]+1;
read(i);
end;
for i:='a' to 'z' do
if a[i]<>0 then write(i,' ',a[i]);
end.
5、输入一段以句号结尾的英文，统计并输出句中各字母出现的次数。假设句中没有大写字母。

const
a=5
x=record;
name1:array[1..a] of string;
number:array[1..a] of string;
cj1,cj2,cj3,cj4:array[1..a] of real;
end;
var
n,e:integer;
b:integer;
s:array[1..a] of real;
pj:array[1..a] of real;
i,j:integer;
zz:array[1..7,1..5] of real;
f:integer;
begin
for b:=1 to 5 do
readln((x.name1[b]),(x.number[b]),(x.cj1[b]),(x.cj2[b]),(x.cj3[b]),(x.cj4[b]));
s[a]:=(x.cj1[b])+(x.cj2[b])+(x.cj3[b])+(x.cj4[b]);
pj:=s[a]/4;
for n:=1 to a do
for e:1 to a-1 do
with x do
if s[e] <s[e+1] then
begin
f:=s[e];
s[e]:=s[e+1];
s[e+1]:=f;
end;
for i:=1 to 7 do
begin
for j:=1 to 5 do
begin
with x do
zz[1,j]:=name;
zz[2,j]:=(cj1[b]);
zz[3,j]:=(cj2[b]);
zz[4,j]:=(cj3[b]);
zz[5,j]:=(cj4[b]);
zz[6,j]:=(zf[b]);
zz[7,j]:=(pj(b));
end;
write(zz[i,j]);
end;
end.
1、利用记录，输入四个姓名，性别，语文、数学、英语、物理的成绩，计算每个人的总分，平均分。最后按照总分从高到低的顺序打印输出所有人的信息。
[有两个错待改－－]
＠朱智语

var
i,n,sum,num,pk,ai:integer;
f,t:string;
function check(a:integer):string;
begin
read(i);
f:='false';
t:='true';
if (i<=2) and (i mod 2<>0 then
check:=f
else if n<i then
ai:=trunc(sqrt(n));
for sum:=1 to ai do
if n mod ai=0 then
a:=i-n;
pk:=trunc(sqrt(a));
for num:=1 to pk do
if a mod num<>0
then check:=t
else check:=f;
write(check);
end.
哥德巴赫猜想[...还是有错]