var n:int64;
z1:longint;
BEGIN
readln(n);
for z1:=2 to trunc(sqrt(n)) do
If n mod z1=0 then
begin
writeln(n div z1);
break;
end;
END.

Block code

var i,n:longint;
begin
read(n);
for i:=2 to round(sqrt(n)) do
if n mod i=0 then
begin
write(n div i);
exit;
end;
end.

这数据也太水了吧

http://user.qzone.qq.com/616674147/infocenter#!app=2&via=QZ.HashRefresh&pos=1376979906

我现在才知道，原来**不要判断素数**。。。。
发一个丑陋的代码：
var i,n:longint;
function prime(p:longint):boolean;var i:longint;
begin
prime:=true;if p=2 then exit(true);
for i:=2 to trunc(sqrt(p)) do if p mod i=0 then exit(false);
end;begin readln(n);
for i:=2 to trunc(sqrt(n)) do
if (prime(i)) and (n mod i=0) then begin write(n div i);break;end;
end.
感觉挺短，但一比较就。。。orz各位犇了

var q,n,p,i:longint;
begin
readln(n);
for i:=2 to n div 2 do
if n mod i=0 then
begin
q:=i;
break;
end;
p:=n div q;
writeln(p);
end.

#include<iostream>
#include<cmath>
int main()
{
using namespace std;
long long n,p;
cin >> n;
p=2;
while(n%p!=0){
p++;
}
p=n/p;
cout << p;
return 0;
}

var n,i:longint;
begin
readln(n);
for i:=2 to n div 2 do
if n mod i=0 then
begin
writeln(n div i);
halt;
end;
end.

i循环需要到n么。。。。水军大战

const
maxn=45000;
var
su:array[1..maxn] of boolean;
n,i,j:longint;
begin
readln(n);
fillchar(su,sizeof(su),true);
su[1]:=false;
for i:=2 to trunc(sqrt(maxn)) do
if su[i] then
for j:=2 to maxn div i do su[i*j]:=false;
for i:=2 to trunc(sqrt(n)) do
if (su[i]=true) and (n mod i=0) then
begin writeln(n div i); break; end;
end.

测试数据 #0: Accepted, time = 0 ms, mem = 736 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 736 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 736 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 732 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 736 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 736 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 732 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 732 KiB, score = 10

测试数据 #8: Accepted, time = 15 ms, mem = 736 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 732 KiB, score = 10

program xx;
var i,n:longint;
begin
read(n);
for i:=2 to n do if n mod i=0 then begin write(n div i);exit; end;
end.
需要那么麻烦么？

VijosEx via JudgeDaemon2/13.6.5.0 via libjudge

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 732 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 732 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 732 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 736 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 732 KiB, score = 10

测试数据 #5: Accepted, time = 3 ms, mem = 736 KiB, score = 10

测试数据 #6: TimeLimitExceeded, time = 1014 ms, mem = 736 KiB, score = 0

测试数据 #7: Accepted, time = 15 ms, mem = 732 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 732 KiB, score = 10

测试数据 #9: Accepted, time = 3 ms, mem = 740 KiB, score = 10

TimeLimitExceeded, time = 1035 ms, mem = 740 KiB, score = 90

怎么会这样

var
a,b,c,d,k:longint;
begin
{assign(input,'prime.in');
reset(input);
assign(output,'prime.out');
rewrite(output);}
read(a);
for b:=2 to a do
if a mod b=0 then
begin
k:=0;
for c:=2 to trunc(sqrt(b)) do
if b mod c=0 then begin k:=1;break;end;
if k=0 then begin write(a div b);break;end;
end;
{close(input);
close(output);}
end.

NOIP普及组 T1 分解质因数（8行代码+解析）

有一个关键点………………

http://user.qzone.qq.com/1057586889/blog/1353143825

program prime;

var

a,b,c,d,k:longint;

begin

{assign(input,'prime.in');

reset(input);

assign(output,'prime.out');

rewrite(output);}

read(a);

for b:=2 to a do

if a mod b=0 then

begin

k:=0;

for c:=2 to trunc(sqrt(b)) do

if b mod c=0 then begin k:=1;break;end;

if k=0 then begin write(a div b);break;end;

end;

{close(input);

close(output);}

end.

源程序

├ 测试数据 01：答案正确... (0ms, 672KB)

├ 测试数据 02：答案正确... (15ms, 672KB)

├ 测试数据 03：答案正确... (15ms, 672KB)

├ 测试数据 04：答案正确... (0ms, 672KB)

├ 测试数据 05：答案正确... (0ms, 672KB)

├ 测试数据 06：答案正确... (15ms, 672KB)

├ 测试数据 07：答案正确... (15ms, 672KB)

├ 测试数据 08：答案正确... (0ms, 672KB)

├ 测试数据 09：答案正确... (15ms, 672KB)

├ 测试数据 10：答案正确... (15ms, 672KB)

---|---|---|---|---|---|---|---|-

Accepted / 100 / 93ms / 672KB

今年的普及组水题

居然不是沙发

此题的样例貌似输错了。。。

我这个Pascal是最省时的，因为我就是无极剑圣，速度天下第一！！！！
var
n,i,max,k:longint;
function f(n:longint):boolean;
var i:longint;
begin
f:=true;
for i:=2 to trunc(sqrt(n)) do
if n mod i=0 then begin f:=false;break;end;
end;
begin
max:=-maxlongint;
read(n);
k:=n;
for i:=1 to trunc(sqrt(k)) do
if (n mod i=0) and (n div i>max) and (f(n div i)=true) then max:=n div i;
write(max);
end.

var
n,i:longint;

begin
read(n);
for i:=2 to n do
if (n mod i=0) then
begin
write(n div i);
exit;
end;
end.

#include <cstdio>
#include <iostream>


using namespace std;

int main(){
    int num;
    cin >> num;
    for (int i = 2; i < num / 2 + 1; i++){
        if (num % i == 0){
            cout << num / i << endl;
            break;
        }
    }
    return 0;
}