Question

求pascal高精度除法程序

即读入a b
输出a/b
保证a是b的正整数倍
fzsdzxccr你在发啥?
我就要高精度除法啊..
参与运算的和结果都是高精度

Answer 1

const n=250; n2=500;
type strn=string[250];
intn22=array [-1..500] of integer;
var a1,f:intn22;
ch1,ch2:strn;
i,n3,n4:integer;
procedure chtoint2(ch:strn;var a:intn22);
{ 将字符串变量转换成整型数组 }
var i,j,t,dot:integer;
begin
{为简单计,将输入数据处理成 ( 0.1, 1 ) 内的数, 用a[0]存放整数部分}
for i:=1 to n4 do a:=0;
j:=0; a[0]:=0; a[-1]:=0;
for i:=1 to length(ch) do
case ch of
'0'..'9': begin j:=j+1; a[j]:=ord(ch)-48 end;
else
end;
t:=0;
repeat t:=t+1 until a[t]>0;
if t>1 then
begin
for i:=1 to j-t+1 do a:=a[i+t-1];
for i:=j-t+2 to j do a:=0;
end;
end;
procedure frmt(var f:intn22; n4:integer);
{ 将数组 f(n4) 规格化 }
var j,fj,fj1:integer;
begin
for j:=n4 downto 1 do
begin
fj:=f[j];
if(fj>9) then
begin fj1:=fj div 10;
f[j-1]:=f[j-1]+fj1;
f[j]:=fj-(fj1*10)
end;
end
end;
procedure multy(var d,e:intn22;var f:intn22; n4:integer);
{ 乘法运算 }
var k,j,dj,nj:integer;
begin
for j:=0 to n4 do f[j]:=0;
for j:=0 to n4 do
begin
dj:=d[j];
if dj>0 then
begin
nj:=n+j;
if nj>n4 then nj:=n4;
for k:=j to nj do
f[k]:=f[k]+dj*e[k-j];
end;
if (j mod 100 =0) then frmt(f,n4);
end;
frmt(f,n4);
f[-1]:=d[-1]+e[-1]
end;
procedure prtc4(var c1:intn22; n3:integer);
{ 打印结果 }
var i:integer;
begin
write(c1[0]:2,'.');
for i:=1 to n3 do write(c1);
end;
procedure initial(a:intn22;var f0:intn22);
{求 f(n2) 的初值, 用 f0 存放}
var d1,d2,e1,e2:real;
k10,i:integer;
begin
k10:=10;
if k10>n3 then k10:=n3;
for i:=1 to n4 do f0:=1;
d1:=0; d2:=0.1;
for i:=1 to k10 do
begin d1:=d1+a*d2; d2:=d2/10 end;
e1:=1/d1;
for i:=1 to k10 do
begin
e2:=e1;
f0[i-1]:=trunc(e2);
e1:=(e1-f0[i-1])*10
end;
f0[-1]:=0;
end;
procedure divid(var a1,f:intn22; n4:integer);
{ a1:分母,f:结果, f1 f2:工作数组 }
var f1,f2: intn22;
i,j,k,kz,t1,t2:integer;
bl:boolean;
begin
initial(a1,f1);
kz:=1;
repeat
inc(kz);
multy(a1,f1,f2,n4);
for i:=1 to n4 do f2:=9-f2;
f2[n2]:=f2[n2]+1; f2[0]:=1-f2[0];
multy(f1,f2,f,n4);
bl:=true;
t1:=1; t2:=n4-2;
if n4>30 then
begin t1:=n4-25; t2:=n4-15; end;
for i:=t1 to t2 do
if f1<>f2 then bl:=false;
if not bl then f1:=f;
until bl or (kz>20);
end;
begin{main}
for i:=1 to n do ch1:=' ';
writeln;
writeln('输入计算位数:(<250)'); readln(n3);
n4:=n3+10;
if n4<n3*1.05 then n4:=trunc(n3*1.05);
writeln('输入分母 a: ( 1>a>=0.1 小数位<250) '); readln(ch1);
chtoint2(ch1,a1);
write('1/'); prtc4(a1,n3); writeln('=');
divid(a1,f,n4);
prtc4(f,n3); writeln;
end.

Answer 2

var a,c,p,d:array[1..1000]of shortint;
i,j,t,k,l:longint;
f,x:boolean;
fin,fout:text;
s:string;
function find:boolean;
begin
find:=false;
for i:=50 downto k+1 do
if (p[i]=p[k])and(d[i]=d[k]) then find:=true;
end;
procedure chu;
begin
t:=0;x:=true;l:=1000;i:=length(s);
while i>0 do begin
t:=t*10+a[i];
if t>=j then begin
c[l]:=t div j;
t:=t mod j;
l:=l-1;
end;
i:=i-1;
end;
k:=50;
if t=0 then exit;
c[l]:=100;l:=l-1;
for i:=1 to 50 do begin p[i]:=0; d[i]:=0; end;
while x and(k>0)and(t<>0) do begin
t:=t*10;
p[k]:=t mod j;
d[k]:=t div j;
t:=t mod j;
if find then x:=false;
k:=k-1;
end;
end;
begin
for i:=1 to 1000 do begin a[i]:=0; c[i]:=0;end;
assign(fin,'l4.in');
reset(fin);
assign(fout,'l4.out');
rewrite(fout);
readln(fin,s);
k:=length(s);
for i:=1 to length(s) do a[i]:=ord(s[length(s)-i+1])-48;
read(fin,j);
chu;
f:=true;i:=1000;
while f and (i>=l+1) do begin
if c[i]=100 then begin
write(fout,'.');
f:=false;
end;
if c[i]<>100 then write(fout,c[i]);
i:=i-1;
end;
if not f then for i:=50 downto k+1 do
write(fout,d[i]);
close(fin);
close(fout);
end.

Answer 3

..............那个一看就知道是在抄袭...其实很简单的...人类是怎么做除法的高精度除法就怎么做....或许你是怕烦吧......

Answer 4

有个问题
是高精度除以高精度吗？
版权归小佑0_0糖所有，盗用者④
lz高精度除法分很多种好不好

Answer 5

这个高精度是什么意思呢？
是参与运算的数是高精度数，还是结果高精度呢？