第一题:计数问题.
program count;
var
n,i,m,t:longint;
begin
assign(input,'count.in');
reset(input);
assign(output,'count.out');
rewrite(output);
readln(n,m);
t:=0;
for i:=1 to n do begin
if (i>=0)and(i<=9) then begin
if i=m then inc(t);
end;
if (i>=10)and(i<=99) then begin
if i div 10 = m then inc(t);
if i mod 10 = m then inc(t);
end;
if (i>=100)and(i<=999) then begin
if i div 100 = m then inc(t);
if i div 10 mod 10 = m then inc(t);
if i mod 10 = m then inc(t);
end;
if (i>=1000)and(i<=9999) then begin
if i div 1000 = m then inc(t);
if i div 100 mod 10 = m then inc(t);
if i div 10 mod 10 = m then inc(t);
if i mod 10 = m then inc(t);
end;
if (i>=10000)and(i<=99999) then begin
if i div 10000 = m then inc(t);
if i div 1000 mod 10 = m then inc(t);
if i div 100 mod 10 = m then inc(t);
if i div 10 mod 10 = m then inc(t);
if i mod 10 = m then inc(t);
end;
if (i>=100000)and(i<=999999) then begin
if i div 100000 = m then inc(t);
if i div 10000 mod 10 = m then inc(t);
if i div 1000 mod 10 = m then inc(t);
if i div 100 mod 10 = m then inc(t);
if i div 10 mod 10 = m then inc(t);
if i mod 10 = m then inc(t);
end;
if (i>=1000000)and(i<=9999999) then begin
if i div 1000000 = m then inc(t);
if i div 100000 mod 10 = m then inc(t);
if i div 10000 mod 10 = m then inc(t);
if i div 1000 mod 10 = m then inc(t);
if i div 100 mod 10 = m then inc(t);
if i div 10 mod 10 = m then inc(t);
if i mod 10 = m then inc(t);
end;
end;
writeln(t);
close(input);
close(output);
end.

二楼提问

var i,n,o,tot:longint;
s,x:string;
begin
assign(input,'count.in');reset(input);
assign(output,'count.out');rewrite(output);
readln(n,x);
delete(x,1,1);
for i:=1 to n do begin
str(i,s);
for o:=1 to length(s) do if s[o]=x then inc(tot);
end;
writeln(tot);
close(input);close(output);
end.
你那是绝对的新手方法

orz LZ

orz

第一题根本就不用这么麻烦，模拟满分！

program expr(input,output);
var n,a,b:longint;ch:char;
begin
assign(input,'expr.in');
reset(input);
assign(output,'expr.out');
rewrite(output);
a:=0;
n:=0;
b:=1;
while not(eoln) do
begin
read(ch);
if ch='+' then
begin
b:=(b*n) mod 10000;
n:=0;
a:=(a+b) mod 10000;
b:=1;
continue;
end;
if ch='*' then
begin
b:=(b*n) mod 10000;
n:=0;
continue;
end;
n:=(n*10+ord(ch)-48) mod 10000;
end;
b:=(b*n) mod 10000;
a:=(a+b) mod 10000;
writeln(a);
writeln();
close(input);
close(output);
end.

program count(input,output);
var n,i,j:longint;x:integer;st:string;total:longint;
begin
assign(input,'count.in');
reset(input);
assign(output,'count.out');
rewrite(output);
readln(n,x);
total:=0;
for i:=1 to n do
begin
str(i,st);
for j:=1 to length(st) do
if st[j]=chr(x+48) then
inc(total);
end;
writeln(total);
writeln();
close(input);
close(output);
end.

　　 　　 　✎　　　灵失空间，我给你留下了这么一句话。

求教第二题解题思路，这里程序看不懂呐~

这位是不会使用循环吧!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

for i:=1 to n do
begin
t:=i;
while i>0 do
begin
if i mod 10=m then inc(tot);
i:=i div 10;
end;
end;
writeln(tot);