请问有没有参考的盲盒智能合约？ | 登链社区 | 技术问答

比如设定有30个A,20个B。 用户随机获得A或B。

struct BoxFactory {
    uint256 id;
    uint8 bunnyId; // A
    string tokenURI;
    uint256 price;
    uint256 limit; // 限制x个，比如A 30
    uint256 minted; // 已经铸造 的个数， 比如A 1个
  }

  // factoryId: BoxFactory
  mapping(uint256 => BoxFactory) public _boxFactories;

function addBoxFactory(
    uint8 bunnyId,
    string memory tokenURI,
    uint256 limit
  ) public onlyOwner returns (uint256) {
    _boxFactoriesId++;

    BoxFactory memory boxFactory;
    boxFactory.id = _boxFactoriesId;
    boxFactory.bunnyId = bunnyId;
    boxFactory.tokenURI = tokenURI;
    boxFactory.price = boxPrice;
    boxFactory.limit = limit;
    _boxFactories[_boxFactoriesId] = boxFactory;
    return _boxFactoriesId;
  }

请问大佬如何实现一个方法，可以公平产出A或B，而且保证A和B的数量不能多也不能少。概率

function buyRandomBox() external payable nonReentrant returns(uint256 tokenId)  {
      // @????? 这里要怎么实现 公平随机决定 产出A还是B
    // uint256 factoryIdAorB??? = Random()????

    BoxFactory storage boxFactory = _boxFactories[factoryId];
    if (boxFactory.limit > 0) {
      require(
        boxFactory.limit.sub(boxFactory.minted) >= 1,
        "Over the limit"
      );
    }
    // 买一个生产一个box
    boxFactory.minted = boxFactory.minted.add(1);

    // 生产
    tokenId = factory.nft.mint(msg.sender, boxFactory.tokenURI, boxFactory.bunnyId);

  }