原力元宇宙MetaForce佛萨奇2.0系统开发
一、什么是原力元宇宙MetaForce？咨询 l8l开259l发3365
原力元宇宙MetaForce是在Polygon马蹄链上部署的一个智能合约，咨询 l8l开259l发3365 Polygon马蹄链，是基于ETH开发的一个独立公链，用于构建和连接与以太坊兼容的区块链网络，智能合约可以直接在马蹄链上部署，百分百开源，百分百去中心化，一旦运行，不可篡改。
Fossage's smart contract address has millions of transactions. Each transaction is transparent from which address to enter. There is no money left in the contract balance, and the project side does not deposit any money, so the project side does not have the problem of encircling money and running away.
用一段比较直白的话解释DApp，那应该是：把我们现在在IOS系统和Android系统开发的App拎出来，放在区块链系统上运行，再结合智能合约，就成了我们所说的DApp了。这么一解释，原来DApp是直接和区块链技术挂钩的，是去中心化的应用。
To put it simply, the fossage metaforce force meta universe project is a smart contract DAPP developed based on horseshoe chain. It is open, transparent and 100% open source. Players go in and out of usdt. There is no routine, and the project party can't tamper with it. The contract is always implemented.
unsigned int APHash(const char *str)
{
unsigned int hash = 0 ;
int i;
for (i= 0 ; *str; i++)
{
if ((i & 1 ) == 0 )
{
hash ^= ((hash << 7 ) ^ (*str++) ^ (hash >> 3 ));
}
else
{
hash ^= (~((hash << 11 ) ^ (*str++) ^ (hash >> 5 )));
}
}
原力元宇宙MetaForce是在Polygon马蹄链上部署的一个智能合约，咨询 l8l开259l发3365 Polygon马蹄链，是基于ETH开发的一个独立公链，用于构建和连接与以太坊兼容的区块链网络，智能合约可以直接在马蹄链上部署，百分百开源，百分百去中心化，一旦运行，不可篡改。
Fossage's smart contract address has millions of transactions. Each transaction is transparent from which address to enter. There is no money left in the contract balance, and the project side does not deposit any money, so the project side does not have the problem of encircling money and running away.
用一段比较直白的话解释DApp，那应该是：把我们现在在IOS系统和Android系统开发的App拎出来，放在区块链系统上运行，再结合智能合约，就成了我们所说的DApp了。这么一解释，原来DApp是直接和区块链技术挂钩的，是去中心化的应用。
To put it simply, the fossage metaforce force meta universe project is a smart contract DAPP developed based on horseshoe chain. It is open, transparent and 100% open source. Players go in and out of usdt. There is no routine, and the project party can't tamper with it. The contract is always implemented.
unsigned int APHash(const char *str)
{
unsigned int hash = 0 ;
int i;
for (i= 0 ; *str; i++)
{
if ((i & 1 ) == 0 )
{
hash ^= ((hash << 7 ) ^ (*str++) ^ (hash >> 3 ));
}
else
{
hash ^= (~((hash << 11 ) ^ (*str++) ^ (hash >> 5 )));
}
}
