Question

此代码的目的：模拟100场CRAPS比赛并记录第一轮赔率，第一轮胜利，第二轮亏损PLUS积分和第二轮胜利PLUS积分。

那些不熟悉CRAPS规则的人;你基本上掷了两个骰子，如果结果是除了总共2,3或12之外的其他任何东西，你就可以再次滚动（你转动的那个数字被保留并加到你的点上）。如果您掷出7或11，则自动获胜。

这是我到目前为止所处的位置：

#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);
}

我问你的是，你给我的选项是如何保留最后打印的那些点？

此外，我觉得我的代码写得更好，冗余更少，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);
}

Answer 1

你宁愿轻易地压缩两次出现的

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

成一个函数：

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

或以单行形式：

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

然后你会写

sumd = rolldice();

Answer 2

将结果放入最后打印的解决方案看起来很合理。如果我理解正确的问题，似乎winp和lostp应该添加sumd2而不是仅仅增加。或者它已经正常工作，我误解了这个问题？

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;
}

C：掷骰子/骰子游戏

2 个答案: