Turbo Pascal编程高手们，高精度减法的程序咋编？？？？？？？？

一个快速入门的秘诀就是“背”，这个“背”字可以大做文章，想当年我就靠这种方法迅速建立起初步信息学知识体系的。

这是《信息学初学者之家》站长、曾多次参加AcM竞赛并获优异成绩的tenshi所发的切身体会。因此，我们要求，下面的算法不仅要读得懂，而且要熟记于心，这样，你才能在程序设计中灵活地运用……

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 { 数的最大长度 }

type

hp=record

len:integer{ 数的长度 }

s:array[1..maxlen] of integer

{ s[1]为最低位，s[len]为最高位 }

end

var

x:array[1..2] of hp

y:hp{ x:输入 y:输出 }

procedure PrintHP(const p:hp)

var i:integer

begin

for i:=p.len downto 1 do write(p.s[i])

end

procedure init {初始化}

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 { 将字符串转换到HP }

x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0')

end

close(input)

end

procedure Plus(a,b:hpvar c:hp){ c:=a+b }

var i,len:integer

begin

fillchar(c,sizeof(c),0)

if a.len>b.len then len:=a.len { 从a,b中取较大长度 }

else len:=b.len

for i:=1 to len do { 从低到高相加 }

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]){ 加1到高位 }

end

end

if c.s[len+1]>0 then inc(len)

c.len:=len

end

procedure out {打印输出}

begin

assign(output,fn_out)

rewrite(output)

PrintHP(y)

writeln

close(output)

end

begin

init

Plus(x[1],x[2],y)

out

end.

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 { 数的最大长度 }

type

hp=record

len:integer{ 数的长度 }

s:array[1..maxlen] of integer

{ s[1]为最低位，s[len]为最高位 }

end

var

x:array[1..2] of hp

y:hp{ x:输入 y:输出 }

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:hpvar 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.

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 { 数的最大长度 }

type

hp=record

len:integer{ 数的长度 }

s:array[1..maxlen] of integer

{ s[1]为最低位，s[len]为最高位 }

end

var

x:array[1..2] of hp

y:hp{ x:输入 y:输出 }

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:hpb:integervar 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:hpvar 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.

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:hpb:integervar c:hpvar 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:hpvar 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:hpvar 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-binc(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.

#include<stdio.h>

#include<time.h>

#include<stdlib.h>

void main()

{

int i,a,b,c,score=0

srand((unsigned)time(NULL))

for(i=1i<=50i++)

{ a=rand()%89+10

b=rand()%89+10

c=rand()%198

if(i<=25)

{

printf("%d+%d=%d\n",a,b,c)

if(c==a+b)

{printf("正确!\n")

score+=2

}

else

printf("错误!\n")

}

else

{

printf("%d-%d=%d\n",a,b,c)

if(c==a-b)

{printf("正确!\n")

score+=2

}

else

printf("错误!\n")

}

}

printf("总分为:%d\n",score)

}