两道编程题目（Pascal语言）

第一题

var

r,x,y:real

begin

readln(r){读入半径}

readln(x,y){读入坐标}

if sqr(x)+sqr(y)=sqr(r) then writeln('yes') {这个是圆心为原点配指的圆的标准方程}

else writeln('no')

end.

第二题

var

s,m,h,i:longint

a:array[1..100000] of longint

begin

readln(a[1]){读培辩配入a1，用数灶迹组读入}

readln(m,h){读入第m项和公差m}

s:=a[1]{用s记录从a1到第m项的乘积}

for i:=2 to m do

begin

a[i]:=a[i-1]+h{计算a[i]的值}

s:=s*a[i]{把a[i]乘到s中去}

end

writeln(s){输出从a1到am的积}

end.

1.输入三个数：A.B.C，输出其中最大和最小的数搏唤局，再输出2A.4B.3C中最大和最小基让的。

函数：

program abc

var i:array[1..3] of longint

a,b,c,x:longint

function fun(var a1,b1,c1:longint):longint

var y:longint

begin

if a1>b1 then begin y:=a1a1:=b1b1:=yend

if a1>c1 then begin y:=a1a1:=c1c1:=yend

if b1>c1 then begin y:=b1b1:=c1c1:=yend

end

begin

readln(a,b,c)

i[1]:=a

i[2]:=b

i[3]:=c

x:=fun(a,b,c)

writeln(a,' ',c)

a:=i[1]

b:=i[2]

c:=i[3]

a:=a*2

b:=b*4

c:=c*3

x:=fun(a,b,c)

writeln(a,' ',c)

end.

过程：

program abc

var i:array[1..3] of longint

a,b,c:longint

procedure maxmin(min,max,y:longint)

begin

for min:=1 to 3-1 do

for max:=min to 3 do

if i[min]>i[max] then begin

y:=i[min]

i[min]:=i[max]

i[max]:=y

end

end

begin

read(a,b,c)

i[1]:=a

i[2]:=b

i[3]:=c

maxmin(i[1],i[2],i[3])

writeln(i[1],' ',i[3])

i[1]:=2*a

i[2]:=4*b

i[3]:=3*c

maxmin(i[1],i[2],i[3])

writeln(i[1],' ',i[3])

end.

2.用递归函数求n！

program JieCheng

var n:longint

s:real

function JC(t:real):real

begin

if t=0 then JC:=1

else JC:=t*JC(t-1)

end

begin

readln(n)

s:=JC(n)

writeln(n,'!=',s:10:10)

end.

3.自定义一个函数D（n）要求把正整数n中间的数链键取出，如果n为正整数则取中间的两位。

program su

var i:array[1..255] of longint

a,b,x:integer

str:string

begin

readln(str)

x:=length(str)

for a:=1 to x do

i[a]:=ord(str[x-a+1])-ord('0')

if x mod 2=0 then begin

b:=x div 2

write('(',str,')=',i[b+1],i[b])

end

else begin

b:=(x+1) div 2

write('(',str,')=',i[b])

end

writeln

end.

4.自定义一个函数把一个给定的十个数的最大和最小值求出。

这道题与第一题相同。

5.自定义一个函数判断输入的a.b是否互为质数。

program hzs

var i1,i2:longint

x:boolean

function hzsfun(k1,k2:longint):boolean

var f:longint

begin

if k1<k2 then begin

f:=k1

k1:=k2

k2:=f

end

if k1 mod k2<>0 then hzsfun:=true

end

begin

read(i1,i2)

x:=hzsfun(i1,i2)

if x=true then writeln('Yes')

else writeln('No')

end.

1:

var a:array[1..1000] of longint

a1,a2:longint

begin

readln(a1){输入总数}

for a1:=1 to a1 do

begin

read(a2){输入每个数字}

inc(a[a2]){这个数字加一}

end

a2:=0{清零}

for a1:=1 to a1 do

if a[a1]>a2 then a2:=a[a1]{看哪个数出现的最多}

write(a2){输出}

end.

2.

var a1,a2,a3:longint

a:array[1..1000] of longint

function ass(a:longint):boolean {判断素数}

var a1:longint

begin

ass:=true

for a1:=2 to trunc(sqrt(a)) do

if a mod a1=0 then ass:=false

end

begin

readln(a1){输入总数}

for a1:=1 to a1 do

read(a[a1]){输入数组}

for a2:=1 to a1 do

for a3:=a2+1 to a1 do

if (ass(a[a2])) and (ass(a[a3])) and (ass((a[a2]+a[a3]) div 2)) and ((a[a2]+a[a3]) mod 2=0) then writeln(a[a2],' ',a[a3],' ',(a[a2]+a[a3]) div 2){判断}

end.

3.

var

a:array[1..10] of longint

a1,a2,a3,a4,a5,a6:longint

begin

for a1:=1 to 10 do read(a[a1]){输入}

for a1:=1 to 7 do

begin

a2:=a[a1]+a[a1+1]+a[a1+2]+a[a1+3]

if a3<a2 then begin a3:=a2a4:=a1end

if (a3>a5) or (a5=0) then begin a5:=a3a6:=a1end

end{判断知山哪一组数大}

writeln('大',a[a4],' ',a[a4+1],' ',a[a4+2],' ',a[a4+3],' ',a4,'~',a4+3)

writeln('小',a[a6].' ',a[a6+1],' ',a[a6+2],' ',a[a6+3],' ',a6,'搭升中~',a6+3)

{输出}

end.

4. {骗分(*^__^*) 嘻嘻……}

不要50，笑祥只要加赏~没事找我

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