题目

<?php 
require __DIR__.'/flag.php';
if (isset($_POST['answer'])){ 
    $number = $_POST['answer']; 
    if (noother_says_correct($number)){ 
        echo $flag; 
    } else { 
        echo "Sorry"; 
    } 
} 

function noother_says_correct($number) 
{ 
    $one = ord('1'); //49
    $nine = ord('9'); //57
    # Check all the input characters! 
    for ($i = 0; $i < strlen($number); $i++) 
    { 
        # Disallow all the digits! 
        $digit = ord($number{$i}); 
        if ( ($digit >= $one) && ($digit <= $nine) ) 
        { 
            # Aha, digit not allowed! 
            return false; 
        } 
    } 
    # Allow the magic number ... 
    return ($number == "3735929054"); //此处原题无括号包裹
} 

highlight_file(__FILE__);
?>

分析

if (isset($_POST['answer']))
POST一发answer

    if (noother_says_correct($number)){ 
        echo $flag;

打印flag的条件是noother_says_correct函数返回true

function noother_says_correct($number) 
{ 
    $one = ord('1'); //49
    $nine = ord('9'); //57
    # Check all the input characters! 
    for ($i = 0; $i < strlen($number); $i++) 
    { 
        # Disallow all the digits! 
        $digit = ord($number{$i}); 
        if ( ($digit >= $one) && ($digit <= $nine) ) 
        { 
            # Aha, digit not allowed! 
            return false; 
        } 
    } 
    # Allow the magic number ... 
    return ($number == "3735929054"); //此处原题无括号包裹
}

该函数对POST的字符串每个字符做检测，如果有一个字符的ASCII码在[49,57]内则返回false，即输入的字符串不能含有1-9数字
返回true的条件是POST的字符串与3735929054等值比较返回true
综合起来后，思路为输入一个不含1-9数字的，与3735929054等值比较返回true的字符串
可以利用弱类型比较和进制转换，找到这么个字符串

知识点

进制

3735929054可以转化为不含1-9的16进制即0xdeadc0de

弱类型比较

challenge2做过介绍，不多说

解法

answer=0xdeadc0de