C：胡扯/骰子游戲

Question

該代碼的目的：模擬CRAPS的100場比賽，並記錄第一輪損失，第一輪勝利，第二輪損失加分和第二輪勝利加分的數量。

你們中那些不熟悉CRAPS規則的人； 基本上可以擲兩個骰子，如果結果不是2、3或12，則可以再次擲骰（擲骰的擲骰數將被保留並加到您的積分中）。 如果您擲出7或11，您將自動獲勝。

這是我到目前為止的位置：

#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);
}

我要問你的是，您給我一些選擇，讓我知道如何保留那些要在最后打印的點？

而且，我覺得我的代碼可以寫得更好，也可以減少重復，建議嗎？

#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

您可以輕松地將兩個事件的發生濃縮

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

進入aa函數：

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

或采用單線形式：

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

那你會寫

sumd = rolldice();

Answer 2

您將結果以int結尾的解決方案看起來很合理。 如果我理解正確的問題，似乎winp和Lostp應該添加sumd2而不是僅僅增加。 還是已經可以正常工作，而我誤讀了這個問題？

您可能還需要查看switch語句：

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;
}