Question

pascal高精度加减乘除？

Answer 1

第二章 高精度计算

2.1 高精度加法
2.2 高精度减法
2.3 高精度乘法

2.4 高精度除法

2.1高精度加法

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 }

procedure PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;

procedure 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;

procedure 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;

procedure main;
begin
Plus(x[1],x[2],y);
end;

procedure 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;

procedure PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;

procedure 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;

procedure 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;

procedure 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;

procedure 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 }

procedure PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;

procedure 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;

procedure 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;

procedure main;
begin
Multiply(x,z,y);
end;

procedure 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 }

procedure PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;

procedure 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;

procedure 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 product 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;

procedure main;
begin
Multiply(x[1],x[2],y);
end;

procedure 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;

procedure PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;

procedure 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;

procedure 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;

procedure main;
begin
Divide(x,z,y,w);
end;

procedure 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 }

procedure PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;

procedure 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;

procedure 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;

procedure 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;

procedure 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;

procedure main;
begin
Divide(x[1],x[2],y,w);
end;

procedure 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.

返回

Answer 2

加:
var st:string;
x,y:array[0..101]of integer;
i,j,l1,l2:integer;
begin
write('x='); readln(st);
l1:=length(st);
for i:=0 to 101 do x[i]:=0;
for i:=l1 downto 1 do
x[l1-i]:=ord(st[i])-ord('0');
write('y='); readln(st);
l2:=length(st);
for i:=0 to 101 do y[i]:=0;
for i:=l2 downto 1 do
y[l2-i]:=ord(st[i])-ord('0');
if l1<l2 then l1:=l2;
for i:=0 to l1 do
begin
x[i]:=x[i]+y[i];
x[i+1]:=x[i+1]+x[i] div 10;
x[i]:=x[i] mod 10;
end;
write('x+y=');
j:=101;
while x[j]=0 do j:=j-1;
for i:=j downto 0 do write(x[i]);
readln;
end.

乘:
var st1,st2:string;
x,y:array[1..200]of integer;
z: array[1..401]of integer; i,j,l1,l2,a,b,c,d:integer;
begin
write('x='); readln(st1);
l1:=length(st1);
for i:=1 to 200 do x[i]:=0;
for i:=l1 downto 1 do
x[l1+1-i]:=ord(st1[i])-ord('0');
write('y='); readln(st2);
l2:=length(st2);
for i:=1 to 200 do y[i]:=0;
for i:=l2 downto 1 do
y[l2+1-i]:=ord(st2[i])-ord('0');
for i:=1 to l1 do
for j:=1 to l2 do
begin
a:=x[i]*y[j];
b:=a div 10;
c:=a mod 10;
d:=i+j-1;{积z数组的元素号}
z[d]:=z[d]+c;
z[d+1]:=z[d+1]+z[d] div 10 + b;{进位}
z[d]:=z[d] mod 10;
end;
a:=l1+l2;;
while z[a]=0 do a:=a-1;
write(st1,’*’,st2,’=‘);
for i:=a downto 1 do write(z[i]);
writeln;
readln;
end.

差:

var st,st1,st2:string;
x,y,z:array[0..200]of byte;
i,k,s1,s2,code:integer;
begin
write('x='); readln(st1);
s1:=length(st1);
write('y='); readln(st2);
s2:=length(st2);
write(st1,’-’,st2,’=‘);
if (s1<s2)or((s1=s2)and(st1<st2)) then
begin
write(‘-’);
st:=st1;st1:=st2;st2:=st;
k:=s1;s1:=s2;s2:=k;
end;
fillchar(x,sizeof(x),0);
fillchar(y,sizeof(y),0);
fillchar(z,sizeof(z),0);
for i:=1 to s1 do val(st1[i],x[s1+1-i],code);
for i:=1 to s2 do val(st2[i],y[s2+1-i],code); for i:= 1 to s1 do
begin
if x[i]-y[i]<0 then{不够减，借位}
begin
x[i+1]:=x[i+1]-1;
x[i]:=x[i] + 10;
end;
z[i]:=x[i]-y[i];
end;
k:=s1;
while z[k]=0 do k:=k-1;
for i:=k downto 1 do write(z[i]);
writeln;
readln;
end.
除:
program HighPrecision3_Multiply1;
const
fn_inp='f:\hp2in.txt';
fn_out='f:\hp5out.txt';
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] 0is the highest position }
end;
var
x,y:hp;
z,w:integer;

procedure PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;

procedure 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;

procedure 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;

procedure main;
begin
Divide(x,z,y,w);
end;

procedure out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
writeln(w);
close(output);
end;

begin
init;
main;
out;
end.
高精度 6 : 除法 2 : 高精度数 / 高精度数
program HighPrecision4_Multiply2;
const
fn_inp='f:\hp2in.txt';
fn_out='f:\hp6out.txt';
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 }

procedure PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;

procedure 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;

procedure 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 c.s[i]:=10+c.s[i]; dec(a.s[i+1])end;

end;
c.len:=len;

end;

function Compare(const a,b:hp):integer;
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;

procedure 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;

procedure 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;

procedure main;
begin
Divide(x[1],x[2],y,w);
end;

procedure out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
PrintHP(w);
writeln;
close(output);
end;

begin
init;
main;
out;
end.