Pascal循环结构题

Question

第一行输入1个整数m
第二行输入70个整数
输出70个整数中最接近m的数（保证只有一个）
给个答案！！！急急急！！！

Answer 1

var m,i,min,min2:longint;
a:array[1..70]of longint;
begin
readln(m);
for i:=1 to 70 do read(a[i]); (读入)
min:=maxlongint;min2:=0;
for i:=1 to 70 do
if abs(a[i]-m)<min then begin min:=abs(a[i]-m);min2:=a[i];end;(寻找)
writeln(min2);(输出)
end.

很简单，学过循环就会了。不会就找我

Answer 2

var m,i,k,k1:longint;
begin
readln(m);
read(k);
for i:=1 to 69 do
begin
read(k1);
if abs(k1-m)<abs(k-m) then
k:=k1;
end;
readln;
writeln(k);
end.

Answer 3

readln(m); k:=10000;
for i:=1 to 70 do
begin
read(t);
if abs(m-t)<k then
begin k:=abs(m-t); ans:=t; end;
end;
writeln(ans);

用差的绝对值来求

Answer 4

program baidu;
var n,m,k:integer;
a:array[1..70] of integer;
begin
read(m);
for n:=1 to 70 do read(a[n]);
k:=a[1];
for n:=2 to 70 do if abs(a[n]-m)<abs(a[n-1]-m) then k:=a[n];
write(k);
end.

Answer 5

var m,i,j,ans:longint;
a:array[1..70] of lonint;
begin
readln(m);
j:=maxlongint;
for i:=1 to 70 do begin read(a[i]);
if abs(a[i]-m)<j then begin
j:=abs(a[i]-m);
ans:=a[i];
end;
end;
writeln(ans);
end.