FreePascal高手看看问题出在哪里？显示错误201

Question

var
ch,kh:array[1..32767] of longint;
i:longint;
num,gao,s,yu:longint;
begin
read(num);
if ((num>=1) and (num<=32767)) then
begin
for i:=1 to 32767 do ch[i]:=0;
i:=1;
while num<>0 do
begin
yu:=num mod 2;
ch[i]:=yu;
num:=num div 2;
i:=i+1;
end;
for i:=1 to 32767 do kh[i]:=ch[32768-i];
gao:=kh[1];
while gao<>1 do
begin
i:=i+1;
gao:=kh[i];
end;
for s:=i to (32768-i) do write(kh[s]);
readln;readln;
end;
end.

Answer 1

很显然这是十进制转二进制的程序：
1、首先，你的‘ch'与'kh'数组的下限不是1，而是0
2、while gao<>1 do 这句话是错误的，二进制的1并非只有末尾有
3、 i:=i+1; gao:=kh[i]; 此时i在循环前的初值是32767，应为’i:=i-1‘
那么，这是我改进的程序：
var
ch,kh:array[0..32767] of longint;i:longint;num,gao,s,yu:longint;
begin read(num);
if ((num>=1) and (num<=32767)) then begin
i:=1;
while num<>0 do begin
yu:=num mod 2;
ch[i]:=yu ;
num:=num div 2;
i:=i+1;
end;
for s:=i-1 downto 1 do write(ch[s]);
readln;
readln;
end;
end.

注意：for s:=i-1 downto 1 是将循环变量s从i-1递减至1

追问

谢谢，试试