Question

Pascal高精度计算，加减乘除

RT，自己能运行成功在提交，并且有部分解释..

Answer 1

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.

Answer 2

var
a,b,c:array[1..200]of 0..9;
n:string;
lena,lenb,lenc,i,x:integer;
begin
readln(n); //输入被加数
lena:=length(n); //被加数放入a数组
for i:=1 to lena do a[lena-i+1]:=ord(n[i])-ord('0');
readln(n); //输入加数
lenb:=length(n); //加数放入b数组 for i:=1 to lenb do b[lenb-i+1]:=ord(n[i])-ord('0');
i:=1;x:=0;
while (i<=lena) or (i<=lenb) do
begin
c[i]:=a[i]+b[i]+x; //两数相加，然后加前进位
x:=c[i] div 10; //x是进位
c[i]:=c[i] mod 10; //保存第i位的值
i:=i+1;
end;
if x>0 then //处理最高进位
begin
lenc:=i;
c[i]:=x;
end
else lenc:=i-1;
for i:=lenc downto 1 do write(c[i]); //输出结果
writeln;
end. 这是加法。。好累，，看下吧，觉得可以的话就继续打

Answer 3

到百度上搜信息初学者之家，里面基本的程序都有