[英]C: Craps / Dice Game
该代码的目的:模拟CRAPS的100场比赛,并记录第一轮损失,第一轮胜利,第二轮损失加分和第二轮胜利加分的数量。
你们中那些不熟悉CRAPS规则的人; 基本上可以掷两个骰子,如果结果不是2、3或12,则可以再次掷骰(掷骰的掷骰数将被保留并加到您的积分中)。 如果您掷出7或11,您将自动获胜。
这是我到目前为止的位置:
#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);
}
我要问你的是,您给我一些选择,让我知道如何保留那些要在最后打印的点?
而且,我觉得我的代码可以写得更好,也可以减少重复,建议吗?
#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);
}
您可以轻松地将两个事件的发生浓缩
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;
}
或采用单线形式:
int rolldice(){
return (rand()%6)+(rand()%6)+2;
}
那你会写
sumd = rolldice();
您将结果以int结尾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。 还是已经可以正常工作,而我误读了这个问题?
您可能还需要查看switch
语句:
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;
}
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