如图，在圆O中，AB=CD，AB、CD的延长线相交于点P，求证：PA=PC

过O作OE垂直AB,OF垂直CD 所以角OEP=角OFP,BE=AE=AB/2,CF=DF=CD/2 因为AB=CD 所以OE=OF,BE=DF 因为OP=OP 所以三角形OEP全等于三角形OFP 所以PE=PF 所以PB=PD 所以PA=PC

我以为多做就是最好的技巧，给你30道

1．请将以下程序段表示的计算公式写出来（假设X的值已给出）

E:=1; A:=1;

FOR N:=1 TO 10 DO

BEGIN

A:=AX/N;

E:=E+A;

ENDFOR;

写出所表示的公式。

2．阅读下列程序段，写出程序段运行后变量X的值。

100 X1=3

X1:=3;

X2:=8;

FOR I:=1 TO 5 DO

BEGIN 循环结构，应用数据轮换方式，求

X:=（X1+X2）2; 两个数和的2倍。

X1:=X2; X2:=X;

END;

WRITELN（‘X=’,X）;

3．阅读下列程序段，写出程序运行后数组元素A1，A2，…，A11中的值。

A[1]:=1;

A[2]:=1; K:=1;

REPEAT

A[K+2]:=1;

FOR I:=K+1 DOWNTO 2 DO

A[I]:=A[I]+A[I-1];

K:=K+1;

UNTIL K>=10;

4． program exp1 (imput,output);

var i,s,max:integer;

a:array [110] of integer;

begin

for i:=1 to 10 do read(a[i]);

max:=a[1]; s:=a[1];

for i:=2 to 10 do

begin

if s<0 then s:=0;

s:=s+a[i];

if s>max then max:=s

end;

writeln(‘max=’,max)

end

输入：-2 13 -1 4 7 8 -1 -18 24 6

输出：max=

输入：8 9 －1 24 6 5 11 15 －28 9

输出：max=

5．program exp2 (input,output);

const n=5;

var i,j,k:integer;

a:array[12n, 12n] of integer;

begin

k:=1;

for i:=1 to 2n-1 do

if i<=n then

if odd(i) then for j:=i downto 1 do begin a[i-j+1,j]:=k; k:=k+1 end

else for j: =1 to i do begin a[i-j+1,j]:=k; k:=k+1; end

else if odd(i) then for j:=n downto i-n+1 do

begin a[i-j+1,j]:=k; k:=k+1; end

else for j:=i-n+1 to n do

begin a[i-j+1,j]:=k; k:=k+1; end;

for i:=1 to n do

begin

for j:=1 to n do write(a[i,j]:3);

writeln

end;

end

6．program exp3 (input,output);

const n=10;

var s,i:integer;

function co(i1:integer):integer;

var j1,s1:integer;

begin

s1:=n; for j1:=(n-1) downto (n-i1+1) do s1:=s1j1 div (n-j1+1);

co:=s1

end;

begin

s:=n+1; for i:=2 to n do s:=s+co(i);

writeln(‘s=’,s);

end

7．program exp4(input,output);

const n=3;

var i,j,s,x:integer;

p:array [0n+1] of integer;

g:array [0100] of integer;

begin

for i := 0 to 100 do g[i]:=0;

p[0]:=0; p[n+1]:=100; for i:= 1 to n do read (p[i]); readln;

for i:= 0 to n do

for j:= i+1 to n+1 do g[abs(p[j]-p[i])]:=g[abs(p[j]-p[i])]+1;

s:=0; for i:=0 to 100 do if g[i]>0 then begin write(i,:4); s:=s+1; end;

writeln; writeln(‘s=’,s); writeln(‘input data:’); readln(x);

writeln(g[x])

end

输入：10 20 65

input data: 10

输出：

8．program excpl;

var x,y,y1,jk,j1,g,e:integcr;

a:array [l20] of 09;

begin

x:=3465; y:=264; jk:=20; for j1:=1 to 20 do a[j1]:=0;

while y＜＞0 do

begin

y1:=y mod 10; y:=y div 10;

while y1＜＞0 do

begin

g:=x;

for e:=jk downto 1 do begin g:=g+a[e]; a[e]:=g mod 10; g:=g div 10 end;

y1:=y1-1

end;

jk:=jk-1

end;

j1:=1; while a[j1]=0 do j1:=j1+1;

for jk:=j1 to 20 do write(a[jk]:4); writeln

end

9．program excp2

var i,j:integer;

a:array [114] of integer;

procedure sw(i1,j1:integer);

var k1:integer;

begin

for k1:=1 to (j1-i1+1) div 2 do

begin

a[i1＋k1-1]:=a[i1＋k1－1]＋a[j1－k1+1];

a[j1－k1＋1]:＝a[i1＋k1－1]－a[j1－k1+1];

a[i1+k1－1]:=a[i1-k1+1]－a[j1－k1+1];

end;

end;

begin

j:=211; for i:=1 to 14 do begin a[i]:=i; j:=j-i end;

sw(1,4); sw( 5,10); sw(11,14); sw(1,14);

for i:=1 to 14 do begin if j mod i=1 then write (a[i]:3); j:=j－a[i]; end;

writeln

end

10．program noi_002;

var i,j,l,n,k,s,t:integer;

b:array [110] of 09;

begin

readln(l,n); s:=l; k:=1; t:=l;

while s<n do begin k:=k+1; t:=tl; s:=s+t end;

s:=s-t; n:=n-s-1; for i:=1 to 10 do b[i]:=0;

j:=11; while n>0 do begin j:=j-1; b[j]:=n mod l; n:=n div l end;

for i:=10-k+1 to 10 do write(chr(ord('a')+b[i]));

end

输入：4 167

输出：

11．program noi_004;

var i,j,j1,j2,p,q:integer;

p1:boolean;

b,c:array [1100] of integer;

begin

readln(q,p); j:=1; p1:=true; b[j]:=q; j1:=0;

while (q>0) and p1 do

begin

j1:=j1+1; c[j1]:=q10 div p; q:=q10-c[j1]p;

if q>0 then begin

j2:=1; while (b[j2]<>q) and (j2<=j) do j2:=j2+1;

if b[j2]=q

then begin

p1:=false; write('0');

for i:=1 to j2-1 do write(c[i]:1);

write('{');

for i:=j2 to j1 do write(c[i]:1);

writeln('}')

end

else begin j:=j+1; b[j]:=q end

end

end;

if q=0 then begin

write('0');

for i:=1 to j1 do write(c[i]:1);

writeln

end;

readln

end

输入 ① 1 8 输出

输入 ② 2 7 输出

12 program chu7_1;

function fun(x：integer)：integer;

begin

if (x=0) or (x=1) then fun:=3

else fun:=x-fun(x-2)

end;

begin

writeln(fun(9));

readln;

end

13program chu7_2;

var i,j,f:integer;

a:array [18] of integer;

begin

for i:=1 to 8 do

begin

f:=i mod 2;

if f=0 then a[i]:=0

else a[i]:=1;

for j:=1 to i do

if f=0 then a[i]:=a[i]+j

else a[i]:=a[i]j

end;

for i:=1 to 8 do write(a[i]:5);

end

14program chu7_3;

var p,q,s,t:integer;

begin

readln(p);

for q:=p+1 to 2p do

begin

t:=0; s:=(pq)mod(q-p);

if s=0 then begin t:=p+q+(pq)div(q-p); write(t:4); end;

end;

readln

end

输入：12

输出：

15prgoram chu7_4;

var n,k,i:integer;

a:array [140] of integer;

procedure find(x:integer);

var s,i1,j1:integer;

p:boolean;

begin

i1:=0; p:=true;

while p do

begin

i1:=i1+1; s:=0;

for j1:=1 to n do if a[j1]>a[i1]then s:=s+1;

if(s=x-1)then begin writeln(a[i1]); p:=false end;

end

end;

begin

readln(n,k);

for i:=1 to n do read(a[i]);

find(k); find(n-k);

end

输入：10 4

12 34 5 65 67 87 7 90 120 13

输出：

16program exp1;

var i,j,k,n,,l0,l1,lk:integer;

a:array [020] of integer;

begin

readln(n,k);

for i:=0 to n-1 do a[i]:=i+1;

a[n]:=a[n-1]; l0:=n-1; lk:=n-1;

for i:=1 to n-1 do

begin

l1:=l0-k; if (l1<0) then l1:=l1+n;

if (l1=lk) then begin a[l0]:=a[n]; lk:=lk-1; a[n]:=a[lk]; l0:=lk end;

else begin a[l0]:=a[l1];l0:=l1; end;

end;

a[l0]:=a[n]; for i:=0 to n-1 do write(a[i]:40); writeln;

end

输入:10 4

输出:

17program exp2;

var n,jr,jw,jb:integer;

ch1:char;

ch:array [120] of char;

begin

readln(n); for i:=1 to n do read(ch[i]); jr:=1; jw:=n; jb:=n;:

while (jr<=jw) do

begin

if (ch[jw]='R')

then begin

ch1:=ch[jr]; ch[jr]:=ch[jw]; ch[jw]:=ch1; jr:=jr+13

end

else if ch[jw]='W' then jw:=jw-1

else begin

ch1:=ch[jw];ch[jw]:=ch[jb];ch[jb]:=ch1;

jw:=jw-1;jb:=jb-1;

end

end;

for i:=1 to n do write(ch[i]);

writeln;

end

输入:10

RBRBWWRBBR

输出:

18pmgram exp3;

var i,j,p,n,q,s:integer;

a:array [120] of integer;

begin

readln(p,n,q);j :=21;

while (n>0)do begin j:=j-1;a[j]:=n mod 10;n:=n div 10; end;

s:=0; for i:=j t0 20 do s:=sp+a[i]; writeln(s); j:=21;

while (s>o)do begin j:=j-1; a[j]:=s mod q; s:=s div q; end;

for i:=j to 20 do write(a[i]);readln;

end

输入:7 3051 8

输出:

19．program programl;

var a,x,y,okl,ok2:integer;

begin

a:=100; x:=l0; y:=20; okl:=5; ok2:=0;

if ((x>y) or ((y<>20) and (okl=0)) and (ok2<>0)) then a:=1

else if ((okl<>0) and (ok2=、0)) then a:=-1 else a:=0;

writeln(a);

end．

20．program Program2;

var a,t:string;

i,j:integer;

begin

a:=morning; j:=l; for i:=2 to 7 do if (a[j]<a[i]) then j:=i;

j:=j-1; for i:=1 to j do write(a[i]);

end．

21．program program3;

var a,b,c,d,sum:longint;

begin

read (a,b,c,d);

a:=a mod 23; b:=b mod 28; c:=c mod 33; sum:=a5544+b 14421+c1288-d;

sum:=sum+21252; sum:=sum mod 21252; if (sum=0) then sum:=21252;

writeln(sum);

end．

输入：283 102 23 320

输出：

22．program program4;

var a:array [05] of integer;

sum,n,max,i,j,k:integer;

cover:array [022000] of Boolean;

begin

read(a[5],a[4],a[3],a[2],a[1],a[0]);

if ((a[5]=0) and (a[3]=0) and (a[1]=0))

then begin a[5]:=a[4]; a[4]:=a[2]; a[3]:=a[0]; a[2]:=0; a[0]:=0; end;

for i:=0 to 5 do if (a[i]>10) then a[i]:=10+(a[i] mod 2);

sum:=0;

for i:=0 to 5 do sum:=sum+a[i](6-i);

if ((sum mod 2)<>0) then begin writeln(‘Can’t be divided’); exit; end;

sum:=sum div 2; max:=0; cover[0]:=True;

for i:=1 to sum2 do cover[i]:=False;

for i:=0 to 5 do;

begin

j:=0;

while (j<a[i]) do

begin

for k:=max downto 0 do

begin if (cover[k]) then cover[k+6-i]:=True; end;

max:=max+6-I; j:=j+1;

end;

end;

if (cover[sum]) then writeln(‘Can be divided’)

else writeln(‘can’t be divided’)；

end．

输入：4 7 9 20 56 48 输入：1000 7 101 20 55 1 输入：2000 5 l 1 0 0

输出： 输出： 输出：

23．program program1;

var

a,b,c,d,e:integer;

begin

a:=79; b:=34; c:=57; d:=0; e:=-1;

if (a<c) or (b>c) then d:=d+e

else if (d+10<e) then d:=e+10

else d:=e-a;

writeln(d);

end

24．program program2;

var

i,j:integer;

str1,str2:string;

begin

str1:='pig-is-stupid'; str2:='clever';

str1[1]:='d'; str1[2]:='o'; i:=8;

for j:=1 to 6 do begin str1[i]:=str2[j]; inc(i); end;

writeln(str1);

end

25．program progam3;

var u:array [03] of integer;

a,b,c,x,y,z:integer;

begin

read(u[0], u[1], u[2], u[3]); a:=u[0]+u[1]+u[2]+u[3]-5;

b:=u[0](u[1]-u[2] div u[3]+8); c:=u[0]u[1] div u[2]u[3];

x:=(a+b+2)3-u[(c+3) mod 4]; y:= (c100-13) div a div (u[b mod 3]5);

if((x+y) mod 2=0) then z:=(a+b+c+x+y) div 2;

z:=(a+b+c–x-y)2; writeln(x+y-z);

end

输入：2 5 7 4

输出： 。

26．program program4;

var c:array [13] of string[200];

s:array [110] of integer;

m,n,i:integer;

procedure numara;

var cod:boolean;

i,j,nr:integer;

begin

for j:=1 to n do begin

nr:=0; cod:=true;

for i:=1 to m do

if c[i,j]='1'

then begin if not cod

then begin cod:=true; inc(s[nr]); nr:=0; end

end

else begin

if cod then begin nr:=1; cod:=false; end

else inc(nr);

end;

if not cod then inc(s[nr]);

end;

end;

begin

readln(m, n); for i:=1 to m do readln(c[i]);

numara; for i:=1 to m do if s[i]<>0 then write(i,' ',s[i],' ');

end

输入：3 10

1110000111

1100001111

1000000011

输出： 。

27．program cup_1;

var n,i,j,k,di,dj:integer;

a:array [120,120] of integer;

begin

n:=7;i:=1;j:=n;k:=1;di:=1;dj:=1;

while (k<=nn) do

begin

a[i,j]:=k; k:=k+1;i:=i+di;j:=j+dj;

if (i=n+1)

then begin i:=n;j:=j-2;di:=-1;dj:=-1 end

else if (j=n+1)

then begin j:=n;di:=-1;dj:=-1 end

else if (j=0)

then begin i:=i+2;j:=1;di:=1;dj:=1 end

else if (i=0) begin j:=1;di:=1;dj:=1 end

end;

for j:=1 to n do write(a[n div 2,j]:4);

writeln

end

28．program cup_2;

var s:string;i,j,m,w:integer;c:char;

a:array [0127] of integer;

begin

for i:=0 to 127 do

begin

c:=chr(i);

if ((’0’<=c) and (c<=’9’))

then a[i]:=1

else

if (((’A’<=c) and (c<=’Z’)) or ((’a’<=c) and (c<=’z’)))

then a[i]:=2

else a[i]:=0

end;

readln(s); m:=length(s);

for i:=1 to m do

begin

j:=ord(s[i]);

if (a[j]>0) then begin write(s[i]);a[j]:=a[j]-1 end

end;

writeln

end

输入：123+1234a+12abc-aaABBABC

输出：

29．program cup_3;

function f(m,n,k:integer):integer;

begin

if ((m=0) and (n=0) and (k=0))

then f:=1

else if ((0<=m) and (m<=n) and (n<=k))

then f:=f(m-1,n,k)+f(m,n-1,k)+f(m,n,k-1)

else f:=0;

end;

begin

writeln(f(3,3,3));

end

30．program cup_4;

var s:string;wc,lcs,i:integer

begin

readln(s);wc:=0;lcs:=1;

for i:=1 to length(s) do

if ((lcs=0) and ((s[i]=’’) or (s[i]=chr(9))))

then lcs:=1

else if ((lcs=0) and (s[i]<>’’) and (s[i]<>chr(9)))

then begin wc:=wc+1;lcs:=0 end;

writeln(wc);

end

输入：This is a test program!

输出：

7 有以下程序

#include <stdioh>

#include <stringh>

main( )

{ char p[20]={‘a’,’b’,’c’,’d’},q[]=”abc”,r[ ]=”abcde”;

strcpy(p+strlen(q),r);

strcat(p,q);

printf(“%d %d\n”,sizeof(p),strlen(p));

}

程序运行后的输出结果是（ C ）。

A）20 9 B) 9 9 C) 20 11 D)11 11

详细解释不懂得不要瞎答

修改后的代码：

void  Print(char arr, int len); //arr改成arr

  void main() 

{

    char  pArray[] = {"Fred","Barrey","Wilma","Betty"};

    int    num = sizeof(pArray) / sizeof(char);       

    printf("Total string numbers = %d\n", num);

    Print(pArray, num);

}

void  Print(char arr, int len)  //arr改成arr

{

    for (int i=0; i<len; i++)

    {

        printf("%s\n", arr[i]);//arr[i]改成arr[i]

    }

}

结果是：3 1

p就是一个包含了三个字符的字符数组，元素个数就是3,；

q是一个字符串数组，编译系统会在字符串尾端自动加上'\0'字符作为结束符，所以，sizeof(q)应该是4， 而q则是字符数组的首地址，代表第一个字符的长度，就是1

以上就是关于如图，在圆O中，AB=CD，AB、CD的延长线相交于点P，求证：PA=PC全部的内容，包括:如图，在圆O中，AB=CD，AB、CD的延长线相交于点P，求证：PA=PC、NOIP读程序的技巧（Pascal）、#include <string.h> #include <stdio.h> main() { char p[20]={'a','b','c','d'},q[]="abc",r[]="abcde";等相关内容解答，如果想了解更多相关内容，可以关注我们，你们的支持是我们更新的动力！