16进制高精度加法计算PASCAL
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发布时间：2022-04-22 01:17
共2个回答
热心网友
时间：2023-09-14 02:54
高精度加法程序如下：
program HighPrecision1_Plus;
const
fn_inp='hp1.inp';
fn_out='hp1.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y:hp; { x:input ; y:output }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Plus(a,b:hp;var c:hp); { c:=a+b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
for i:=1 to len do { plus from low to high }
begin
inc(c.s[i],a.s[i]+b.s[i]);
if c.s[i]>=10 then
begin
dec(c.s[i],10);
inc(c.s[i+1]); { add 1 to a higher position }
end;
end;
if c.s[len+1]>0 then inc(len);
c.len:=len;
end;
procere main;
begin
Plus(x[1],x[2],y);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2. 2 高精度减法
高精度减法程序如下：
program HighPrecision2_Subtract;
const
fn_inp='hp2.inp';
fn_out='hp2.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y:hp; { x:input ; y:output }
positive:boolean;
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Subtract(a,b:hp;var c:hp); { c:=a-b, suppose a>=b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
for i:=1 to len do { subtract from low to high }
begin
inc(c.s[i],a.s[i]-b.s[i]);
if c.s[i]<0 then
begin
inc(c.s[i],10);
dec(c.s[i+1]); { add 1 to a higher position }
end;
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
function Compare(const a,b:hp):integer;
{
1 if a>b
0 if a=b
-1 if a<b
}
var len:integer;
begin
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
while(len>0) and (a.s[len]=b.s[len]) do dec(len);
{ find a position which have a different digit }
if len=0 then compare:=0 { no difference }
else compare:=a.s[len]-b.s[len];
end;
procere main;
begin
if Compare(x[1],x[2])<0 then positive:=false
else positive:=true;
if positive then Subtract(x[1],x[2],y)
else Subtract(x[2],x[1],y);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
if not positive then write('-');
PrintHP(y);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2.3高精度乘法
1.高精度乘单精度(1位数)
程序如下:
program HighPrecision3_Multiply1;
const
fn_inp='hp3.inp';
fn_out='hp3.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x,y:hp; { x:input hp ; y:output }
z:integer; { z:input lp }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
i:integer;
begin
assign(input,fn_inp);
reset(input);
readln(st);
x.len:=length(st);
for i:=1 to x.len do { change string to HP }
x.s[i]:=ord(st[x.len+1-i])-ord('0');
readln(z);
close(input);
end;
procere Multiply(a:hp;b:integer;var c:hp); { c:=a*b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a.len;
for i:=1 to len do
begin
inc(c.s[i],a.s[i]*b);
inc(c.s[i+1],c.s[i] div 10);
c.s[i]:=c.s[i] mod 10;
end;
inc(len);
while(c.s[len]>=10) do
begin
inc(c.s[len+1],c.s[len] div 10);
c.s[len]=c.s[len] mod 10;
inc(len);
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Multiply(x,z,y);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2.高精度乘一个整型数据(integer)
只需要将上述程序的hp类型定义如下即可:
type
hp=record
len:integer { length of the number }
s:array[1..maxlen] of longint
{ s[1] is the lowest position
s[len] is the highest position }
end;
3.高精度乘高精度
程序如下:
program HighPrecision4_Multiply2;
const
fn_inp='hp4.inp';
fn_out='hp4.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y:hp; { x:input ; y:output }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Multiply(a,b:hp;var c:hp); { c:=a+b }
var i,j,len:integer;
begin
fillchar(c,sizeof(c),0);
for i:=1 to a.len do
for j:=1 to b.len do
begin
inc(c.s[i+j-1],a.s[i]*b.s[j]);
inc(c.s[i+j],c.s[i+j-1] div 10);
c.s[i+j-1]:=c.s[i+j-1] mod 10;
end;
len:=a.len+b.len+1;
{
the proct of a number with i digits and a number with j digits
can only have at most i+j+1 digits
}
while(len>1)and(c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Multiply(x[1],x[2],y);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2.4 高精度除法
1.高精度除以整型数据(integer);
程序如下:
program HighPrecision3_Multiply1;
const
fn_inp='hp5.inp';
fn_out='hp5.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x,y:hp;
z,w:integer;
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
i:integer;
begin
assign(input,fn_inp);
reset(input);
readln(st);
x.len:=length(st);
for i:=1 to x.len do { change string to HP }
x.s[i]:=ord(st[x.len+1-i])-ord('0');
readln(z);
close(input);
end;
procere Divide(a:hp;b:integer;var c:hp;var d:integer);
{ c:=a div b ; d:=a mod b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a.len;
d:=0;
for i:=len downto 1 do { from high to low }
begin
d:=d*10+a.s[i];
c.s[i]:=d div b;
d:=d mod b;
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Divide(x,z,y,w);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
writeln(w);
close(output);
end;
begin
init;
main;
out;
end.
2.高精度除以高精度
程序如下:
program HighPrecision4_Multiply2;
const
fn_inp='hp6.inp';
fn_out='hp6.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y,w:hp; { x:input ; y:output }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Subtract(a,b:hp;var c:hp); { c:=a-b, suppose a>=b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
for i:=1 to len do { subtract from low to high }
begin
inc(c.s[i],a.s[i]-b.s[i]);
if c.s[i]<0 then
begin
inc(c.s[i],10);
dec(c.s[i+1]); { add 1 to a higher position }
end;
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
function Compare(const a,b:hp):integer;
{
1 if a>b
0 if a=b
-1 if a<b
}
var len:integer;
begin
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
while(len>0) and (a.s[len]=b.s[len]) do dec(len);
{ find a position which have a different digit }
if len=0 then compare:=0 { no difference }
else compare:=a.s[len]-b.s[len];
end;
procere Multiply10(var a:hp); { a:=a*10 }
var i:Integer;
begin
for i:=a.len downto 1 do
a.s[i+1]:=a.s[i];
a.s[1]:=0;
inc(a.len);
while(a.len>1) and (a.s[a.len]=0) do dec(a.len);
end;
procere Divide(a,b:hp;var c,d:hp); { c:=a div b ; d:=a mod b }
var i,j,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a.len;
fillchar(d,sizeof(d),0);
d.len:=1;
for i:=len downto 1 do
begin
Multiply10(d);
d.s[1]:=a.s[i]; { d:=d*10+a.s[i] }
{ c.s[i]:=d div b ; d:=d mod b; }
{ while(d>=b) do begin d:=d-b;inc(c.s[i]) end }
while(compare(d,b)>=0) do
begin
Subtract(d,b,d);
inc(c.s[i]);
end;
end;
while(len>1)and(c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Divide(x[1],x[2],y,w);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
PrintHP(w);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
练习：
求n!(n<10000)的最后一位不为0的数字.
热心网友
时间：2023-09-14 02:54
program Eugene;
var i,j,xx:longint;
ch:char;
num:string;
begin
readln(num); xx:=0;
for i:=1 to length(num) do
begin
case num[i] of
'0'..'9': j:=ord(num[i])-48;
'A'..'F': j:=ord(num[i])-55;
end;
xx:=xx*16+j;
end;
num:='';
while xx<>0 do
begin
case xx mod 10 of
0..9: ch:=chr(xx mod 10+48);
10..15: ch:=chr(xx mod 10+55);
end;
num:=ch+num; xx:=xx div 10;
end;
writeln(num);
end.
输入一个十六进制数num即可……
自己编的~希望对你有帮助哈……O(∩_∩)O~
（注：FP上运行OK~）
具体格式见下面：
参考资料：http://hi.baidu.com/royaleugene/blog/item/9d3aac3ee75a13ca7c1e71ee.html
热心网友
时间：2023-09-14 02:54
高精度加法程序如下：
program HighPrecision1_Plus;
const
fn_inp='hp1.inp';
fn_out='hp1.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y:hp; { x:input ; y:output }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Plus(a,b:hp;var c:hp); { c:=a+b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
for i:=1 to len do { plus from low to high }
begin
inc(c.s[i],a.s[i]+b.s[i]);
if c.s[i]>=10 then
begin
dec(c.s[i],10);
inc(c.s[i+1]); { add 1 to a higher position }
end;
end;
if c.s[len+1]>0 then inc(len);
c.len:=len;
end;
procere main;
begin
Plus(x[1],x[2],y);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2. 2 高精度减法
高精度减法程序如下：
program HighPrecision2_Subtract;
const
fn_inp='hp2.inp';
fn_out='hp2.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y:hp; { x:input ; y:output }
positive:boolean;
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Subtract(a,b:hp;var c:hp); { c:=a-b, suppose a>=b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
for i:=1 to len do { subtract from low to high }
begin
inc(c.s[i],a.s[i]-b.s[i]);
if c.s[i]<0 then
begin
inc(c.s[i],10);
dec(c.s[i+1]); { add 1 to a higher position }
end;
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
function Compare(const a,b:hp):integer;
{
1 if a>b
0 if a=b
-1 if a<b
}
var len:integer;
begin
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
while(len>0) and (a.s[len]=b.s[len]) do dec(len);
{ find a position which have a different digit }
if len=0 then compare:=0 { no difference }
else compare:=a.s[len]-b.s[len];
end;
procere main;
begin
if Compare(x[1],x[2])<0 then positive:=false
else positive:=true;
if positive then Subtract(x[1],x[2],y)
else Subtract(x[2],x[1],y);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
if not positive then write('-');
PrintHP(y);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2.3高精度乘法
1.高精度乘单精度(1位数)
程序如下:
program HighPrecision3_Multiply1;
const
fn_inp='hp3.inp';
fn_out='hp3.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x,y:hp; { x:input hp ; y:output }
z:integer; { z:input lp }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
i:integer;
begin
assign(input,fn_inp);
reset(input);
readln(st);
x.len:=length(st);
for i:=1 to x.len do { change string to HP }
x.s[i]:=ord(st[x.len+1-i])-ord('0');
readln(z);
close(input);
end;
procere Multiply(a:hp;b:integer;var c:hp); { c:=a*b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a.len;
for i:=1 to len do
begin
inc(c.s[i],a.s[i]*b);
inc(c.s[i+1],c.s[i] div 10);
c.s[i]:=c.s[i] mod 10;
end;
inc(len);
while(c.s[len]>=10) do
begin
inc(c.s[len+1],c.s[len] div 10);
c.s[len]=c.s[len] mod 10;
inc(len);
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Multiply(x,z,y);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2.高精度乘一个整型数据(integer)
只需要将上述程序的hp类型定义如下即可:
type
hp=record
len:integer { length of the number }
s:array[1..maxlen] of longint
{ s[1] is the lowest position
s[len] is the highest position }
end;
3.高精度乘高精度
程序如下:
program HighPrecision4_Multiply2;
const
fn_inp='hp4.inp';
fn_out='hp4.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y:hp; { x:input ; y:output }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Multiply(a,b:hp;var c:hp); { c:=a+b }
var i,j,len:integer;
begin
fillchar(c,sizeof(c),0);
for i:=1 to a.len do
for j:=1 to b.len do
begin
inc(c.s[i+j-1],a.s[i]*b.s[j]);
inc(c.s[i+j],c.s[i+j-1] div 10);
c.s[i+j-1]:=c.s[i+j-1] mod 10;
end;
len:=a.len+b.len+1;
{
the proct of a number with i digits and a number with j digits
can only have at most i+j+1 digits
}
while(len>1)and(c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Multiply(x[1],x[2],y);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2.4 高精度除法
1.高精度除以整型数据(integer);
程序如下:
program HighPrecision3_Multiply1;
const
fn_inp='hp5.inp';
fn_out='hp5.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x,y:hp;
z,w:integer;
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
i:integer;
begin
assign(input,fn_inp);
reset(input);
readln(st);
x.len:=length(st);
for i:=1 to x.len do { change string to HP }
x.s[i]:=ord(st[x.len+1-i])-ord('0');
readln(z);
close(input);
end;
procere Divide(a:hp;b:integer;var c:hp;var d:integer);
{ c:=a div b ; d:=a mod b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a.len;
d:=0;
for i:=len downto 1 do { from high to low }
begin
d:=d*10+a.s[i];
c.s[i]:=d div b;
d:=d mod b;
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Divide(x,z,y,w);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
writeln(w);
close(output);
end;
begin
init;
main;
out;
end.
2.高精度除以高精度
程序如下:
program HighPrecision4_Multiply2;
const
fn_inp='hp6.inp';
fn_out='hp6.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y,w:hp; { x:input ; y:output }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Subtract(a,b:hp;var c:hp); { c:=a-b, suppose a>=b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
for i:=1 to len do { subtract from low to high }
begin
inc(c.s[i],a.s[i]-b.s[i]);
if c.s[i]<0 then
begin
inc(c.s[i],10);
dec(c.s[i+1]); { add 1 to a higher position }
end;
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
function Compare(const a,b:hp):integer;
{
1 if a>b
0 if a=b
-1 if a<b
}
var len:integer;
begin
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
while(len>0) and (a.s[len]=b.s[len]) do dec(len);
{ find a position which have a different digit }
if len=0 then compare:=0 { no difference }
else compare:=a.s[len]-b.s[len];
end;
procere Multiply10(var a:hp); { a:=a*10 }
var i:Integer;
begin
for i:=a.len downto 1 do
a.s[i+1]:=a.s[i];
a.s[1]:=0;
inc(a.len);
while(a.len>1) and (a.s[a.len]=0) do dec(a.len);
end;
procere Divide(a,b:hp;var c,d:hp); { c:=a div b ; d:=a mod b }
var i,j,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a.len;
fillchar(d,sizeof(d),0);
d.len:=1;
for i:=len downto 1 do
begin
Multiply10(d);
d.s[1]:=a.s[i]; { d:=d*10+a.s[i] }
{ c.s[i]:=d div b ; d:=d mod b; }
{ while(d>=b) do begin d:=d-b;inc(c.s[i]) end }
while(compare(d,b)>=0) do
begin
Subtract(d,b,d);
inc(c.s[i]);
end;
end;
while(len>1)and(c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Divide(x[1],x[2],y,w);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
PrintHP(w);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
练习：
求n!(n<10000)的最后一位不为0的数字.
热心网友
时间：2023-09-14 02:54
program Eugene;
var i,j,xx:longint;
ch:char;
num:string;
begin
readln(num); xx:=0;
for i:=1 to length(num) do
begin
case num[i] of
'0'..'9': j:=ord(num[i])-48;
'A'..'F': j:=ord(num[i])-55;
end;
xx:=xx*16+j;
end;
num:='';
while xx<>0 do
begin
case xx mod 10 of
0..9: ch:=chr(xx mod 10+48);
10..15: ch:=chr(xx mod 10+55);
end;
num:=ch+num; xx:=xx div 10;
end;
writeln(num);
end.
输入一个十六进制数num即可……
自己编的~希望对你有帮助哈……O(∩_∩)O~
（注：FP上运行OK~）
具体格式见下面：
参考资料：http://hi.baidu.com/royaleugene/blog/item/9d3aac3ee75a13ca7c1e71ee.html
