0

此代码的目的：模拟 100 场 CRAPS 游戏，并记录第一轮输、第一轮获胜、第二轮输 PLUS 点数和第二轮获胜 PLUS 点数。

那些不熟悉 CRAPS 规则的人；你基本上掷了两个骰子，如果结果不是总数为 2、3 或 12，你可以再次掷骰子（你在该回合中掷出的数字被保留并添加到你的分数中）。如果您掷出 7 或 11，您将自动获胜。

这是我目前所处的位置：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main ()
{
int i,d1,d2,sumd,sumd2;
double winf = 0, lostf = 0, winp = 0, lostp = 0;
printf("This program will simulate the game of craps for 100 times.\n");

for (i=0; i<100; i++) {
    d1 = rand()%6+1;
    d2 = rand()%6+1;
    sumd = d1 + d2;

    if (sumd==7 || sumd==11) {
        printf("You rolled a 7 or an 11, you win.\n");
        winf++;
    }
    if (sumd==2 || sumd==3 || sumd==12) {
        printf("You rolled a 12, a 3, or a 2, you lose.\n");
        lostf++;
    }
    if (sumd==4 || sumd==5 || sumd==6 || sumd==8 || sumd==9 || sumd==10) {
        while (1) {
            d1 = rand()%6+1;
            d2 = rand()%6+1;
            sumd2 = d1 + d2;

            if (sumd2==sumd){ 
                printf("You rolled your points, you win.\n");
                winp++;
            break;}
            if (sumd==7){ 
                printf("You rolled a 7, you lose.\n");
                lostp++;
            break;}
        }
    }
}

printf("First roll wins: %lf, First roll loses: %lf, Second roll wins: %lf, Second roll loses: %lf. ", winf, lostf, winp, lostp);
}

我对你的要求是，你给我一些选项，让我可以选择如何保留这些点，以便在最后打印？

此外，我觉得我的代码可以写得更好，更少冗余，建议？

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main ()
{
int i,d1,d2,sumd,sumd2;
double winf = 0, lostf = 0, winp = 0, lostp = 0;

printf("This program will simulate the game of craps for 100 times. Press any key to continue.\n");
//getchar();

for (i=0; i<100; i++) {
    d1 = rand()%6+1;
    d2 = rand()%6+1;
    sumd = d1 + d2;

switch(sumd){
    case 7:
    case 11:
        printf("You rolled %d, you win.\n", sumd);
        winf++;
        break;
    case 2:
    case 3:
    case 12:
        printf("You rolled %d, you lose.\n", sumd);
        lostf++;
        break;
    default:
        while (1) {
            d1 = rand()%6+1;
            d2 = rand()%6+1;
            sumd2 = d1 + d2;

            if (sumd2==sumd){ 
                printf("You rolled your points(%d), you win.\n",sumd);
                winp++;
            break;}
            if (sumd2==7){ 
                printf("You rolled a 7, you lose.\n");
                lostp++;
            break;}
        }
}

}
printf("First roll wins: %lf, First roll loses: %lf, Second roll wins: %lf, Second roll loses: %lf. \n", winf, lostf, winp, lostp);
}
4

2 回答 2

2

你可以很容易地浓缩这两种情况

d1 = rand()%6+1;
d2 = rand()%6+1;
sumd2 = d1 + d2;

进入aa函数：

int rolldice(){
    int d1,d2;
    d1 = rand()%6+1;
    d2 = rand()%6+1;
    return d1 + d2;
}

或以单线形式：

int rolldice(){
    return (rand()%6)+(rand()%6)+2;
}

然后你会写

sumd = rolldice();
于 2012-02-08T08:35:55.747 回答
1

您将结果放入 ints 以在最后打印的解决方案看起来很合理。如果我正确理解了这个问题，似乎 winp 和 lostp 应该添加 sumd2 而不是仅仅增加。还是它已经可以正常工作并且我误读了这个问题？

您可能还想查看switch声明：

switch(sumd){
    case 7:
    case 11:
        //existing code goes here
        break;

    case 2:
    case 3:
    case 12:
        //more existing code
        break;

    default:
        //code for games that don't end on the first turn
        break;
}
于 2012-02-08T07:58:31.723 回答