Pascal输入一串字符,以?结束，分别统计其中数字（0,1,2，....9）出现的次数

程序很简单，仔细看吧：
var
a:array ['0'..'9'] of longint;
b:array ['0'..'9'] of char;
c,i,j,tt:char;
t:longint;
begin
//初始化
for i:='0' to '9' do
b[i]:=i;
//读取字符
read(c);
while c<>'?' do
begin
if (c>='0')and(c<='9') then
inc(a[c]);
read(c);
end;
//冒泡排序
for i:='0' to '8' do
for j:=chr(ord(i)+1) to '9' do
if a[i]<a[j] then
begin
t:=a[i];
a[i]:=a[j];
a[j]:=t;
tt:=b[i];
b[i]:=b[j];
b[j]:=tt;
end;
//输出统计结果
for i:='0' to '9' do
writeln(b[i],' ',a[i]);
end.本回答被提问者采纳

var ch:char;
a:array['0'..'9'] of integer;
begin
while true do
begin
read(ch);
if ch='?' then break;
inc(a[ch]);
end;

for ch:='0' to '9' do
write(a[ch],' ');
end.

var
a:array[0..9] of integer;
b:array[0..9] of boolean;
c:char;
i,max,t:integer;
begin
repeat

read(c);
if (c<>'?') and ('0'<c) and (c<'9') then
begin
a[ord(c)-48]:=1+a[ord(c)-48];
end;
until c='?';

for i:=0 to 9 do
begin
b[i]:=true;
end;

max:=0;
t:=0;

repeat
max:=0;
for i:=0 to 9 do
if (a[i]>max) and (b[i]) then max:=a[i];

for i:=0 to 9 do
if a[i]=max then
begin
writeln(i,' ',a[i]);
b[i]:=false;
inc(t);
end;
until t=10;
end.

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