我已经定义了一种计算获胜几率的方法,并返回支付给玩家的金额。我最初写这是完成此任务最明显的方法,但我真的想知道重构这个的最佳方法,以最大限度地减少计算机的工作量。这是我的功能:
int getPay(int winningOdds, int payOut)
{
winningOdds = rand() % MAXIMUM_ODDS + MINIMUM_ODDS;
if(winningOdds == 1)
{
payOut = 4000;
printf("Jackpot! - %d\n", payOut);
}
else if(winningOdds <= 2)
{
payOut = 1000;
printf("Coins won - %d\n", payOut);
}
else if(winningOdds <= 3)
{
payOut = 500;
printf("Coins won - %d\n", payOut);
}
else if(winningOdds <= 40)
{
payOut = 100;
printf("Coins won - %d\n", payOut);
}
else if(winningOdds <= 50)
{
payOut = 50;
printf("Coins won - %d\n", payOut);
}
else if(winningOdds <= 75)
{
payOut = 25;
printf("Coins won - %d\n", payOut);
}
else if(winningOdds <= 1000)
{
payOut = 10;
printf("Coins won - %d\n", payOut);
}
else if(winningOdds <= 3000)
{
payOut = 5;
printf("Coins won - %d\n", payOut);
}
else if(winningOdds <= 5000)
{
payOut = 3;
printf("Coins won - %d\n", payOut);
}
else if(winningOdds <= 10000)
{
payOut = 1;
printf("Coins won - %d\n", payOut);
}
return payOut;
}
答案 0 :(得分:2)
尝试这种风格,您需要填写Payout
struct Payout{
int winningOdds, payOut;
};
int getPay(int winningOdds, int payOut)
{
Payout payout[] = {{1,4000},{2,1000}....};
int numElem = sizeof(payout)/sizeof(payout[0]);
winningOdds = rand() % MAXIMUM_ODDS + MINIMUM_ODDS;
for(int i=0;i<numElem ;i++){
if(winningOdds <=payout[i].winningOdds){
payOut = payout[i].payout;
break;
}
}
printf("Coins won - %d\n", payOut);
}