p2392.pas
来自「高手写的所有acm例程 在acm.zju.edu.cn 上的题目的例程」· PAS 代码 · 共 116 行
PAS
116 行
PROGRAM p2392;
TYPE
AnsType =array[0..9]of Qword;
VAR
order,part :array[0..10]of Longint;
PROCEDURE Print(var t:AnsType);
var
i :Integer;
begin
for i:=0 to 9 do
begin
write(t[i]);
if I<9 then write(' ');
end;
writeln;
end;
FUNCTION Long(n:Longint):Integer;
begin
long:=trunc(ln(n)/ln(10))+1;
end;
PROCEDURE Add(var a,b:AnsType);
var
i :Integer;
begin
for i:=0 to 9 do a[i]:=a[i] + b[i];
end;
FUNCTION f(n:Longint):AnsType;
var
l,p,first,i :Integer;
fa,a,b :AnsType;
begin
if n=0 then
begin
fillchar(fa,sizeof(fa),0);
f:=fa;
exit;
end; //arim
l:=long(n);
first:=n div order[l];
a:=f(order[l]-1);
fa:=a;
p:=l-1;
while (p>0)and(n div order[p] mod 10=0) do dec(p);
if (p<>0) then
begin
fa[0]:=fa[0]+(n mod order[p+1]) * (l-p-1);
fa[first]:=fa[first]+(n mod order[p+1]);
b:=f(n mod order[p+1]);
Add(fa,b);
fa[0]:=fa[0]+part[p];
end;
a[0]:=a[0]+part[l-1];
for i:=1 to first-1 do
fa[i]:=fa[i]+order[l];
fa[first]:=fa[first]+1;
fa[0]:=fa[0]+(l-1)*first;
for i:=1 to first-1 do add(fa,a);
f:=fa;
end;
PROCEDURE MakeOrderAndPart;
var
i :Integer;
begin
order[0]:=1;
part[0]:=0;
order[1]:=1;
part[1]:=0;
for i:=2 to 10 do
begin
order[i]:=order[i-1]*10;
part[i]:=part[i-1] + order[i]-1;
end;
end;
PROCEDURE main;
var
temp1,temp2 :AnsType;
a,b,i,temp :Longint;
begin
while true do
begin
readln(a,b);
if (a=0)and(b=0) then break;
if a>b then
begin
temp:=a;
a:=b;
b:=temp;
end;
temp1:=f(a-1);
temp2:=f(b);
for i:=0 to 9 do temp2[i]:=temp2[i]-temp1[i];
print(temp2);
end;
end;
BEGIN
//assign(input,'p.in');
//reset(input);
MakeOrderAndPart;
main;
END.
⌨️ 快捷键说明
复制代码Ctrl + C
搜索代码Ctrl + F
全屏模式F11
增大字号Ctrl + =
减小字号Ctrl + -
显示快捷键?