Guess the secret number

1. 题目

1.1 Putting the answer in the code makes things a little too easy.

This time I’ve only stored the hash of the number. Good luck reversing a cryptographic hash!
1.2 题目代码：

// SPDX-License-Identifier: MIT
pragma solidity ^0.4.21;

contract GuessTheSecretNumberChallenge {
    bytes32 answerHash = 0xdb81b4d58595fbbbb592d3661a34cdca14d7ab379441400cbfa1b78bc447c365;

    function GuessTheSecretNumberChallenge() public payable {
        require(msg.value == 1 ether);
    }
    
    function isComplete() public view returns (bool) {
        return address(this).balance == 0;
    }

    function guess(uint8 n) public payable {
        require(msg.value == 1 ether);

        if (keccak256(n) == answerHash) {
            msg.sender.transfer(2 ether);
        }
    }
}

2. 分析

2.1 这是我第一次感觉自己能想出来的题目了。。。。
2.2 反正 uint8 的范围是 0-256 嘛，直接循环一个一个蛮力法，一个一个试出来
2.3 写一个Hack 合约来计算，哈希值为：0xdb81b4d58595fbbbb592d3661a34cdca14d7ab379441400cbfa1b78bc447c365 的数字
Hack 合约代码：

contract Hack {
    uint8 public number;
    bytes32 answer = 0xdb81b4d58595fbbbb592d3661a34cdca14d7ab379441400cbfa1b78bc447c365;

    function f() public returns(uint8) {
        for (uint8 i = 0; i <= 256; i++) {
            if (keccak256(i) == answer) {
                number = i;
                break;
            }
        }
        return number;
    }
}

运行求出来的结果是：170

3. 解题

3.1 部署合约
3.2 将 170 作为参数，调用guss() 函数
3.3 同时前两步操作都需要将 msg.value 设置为 1 ether
![image-20240412144923642](Guess the secret number/image-20240412144923642.png)