猜数字游戏v2.0[C语言][自制]

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和室友做的第二个版本。具体就是判断四位数字是否相同，然后给出哪个位相同，哪个位不同XD。

具体思路：

用随机函数产生四位数字，然后一位位比较。这个是个so easy 版本，没有猜测次数限制XD

效果图：

代码如下：

/*GuessNumberGame2*/#include<stdio.h>#include<string.h>#include<stdlib.h>#include<time.h>#pragma warning(disable:4996)char random_number[4];void show_line(void)   //产生分界线{int i;for (i = 0; i < 80; i++)putchar('=');}void creat(void)  //产生四位数{srand((unsigned)time(NULL));for (int i = 0; i < 4; i++)    {random_number[i] = '0' + rand() % 10;if (i == 0 && random_number[i] == '0') i--; //防止开头是0    }}void warning(void){printf("Please enter 4 positive numbers.eg:1123,1234,9943.\n\n");}int judge(char s[4])     //判断是否是四位数{int i;if (s[0] == '0')return 0;for (i = 0; i < 4; i++)if (s[i]<'0' || s[i]>'9')return 0;return 1;}int main(void){int JUDGE;char guessnumber[5];printf("Guess Number Game 2 [Version 0.6]\n""Copyright <c> 2014  632 Technical Team. All Right Reserved.\n\n");show_line();printf("这是一个猜数字游戏，规则如下：\n"           "系统将随机产生一个四位数，请您进行猜测，我们会将"   "您输入的数字进行匹配。\n");creat(); //产生四位数beg:printf("Please enter four integer numbers:\n");printf("依次输入四位数的每一位数\n");gets(guessnumber);JUDGE=judge(guessnumber);   //判断是否是四位数while (JUDGE == 1){printf("The number you write is:\n");for (int i = 0; i < 4; i++)printf("%c", guessnumber[i]);        printf("\n");int f = 0, cnt = 0;while (f < 4){if (random_number[f] == guessnumber[f]){printf("第%d位等\n", f + 1);++cnt; //统计是否都是正确的}else{printf("第%d位不等\n", f + 1);}f++;}if (cnt == 4)        {            printf("你猜对啦~\n");            goto ed;        }show_line();printf("Please enter four integer numbers:\n");printf("依次输入四位数的每一位数例如:\n");gets(guessnumber);JUDGE=judge(guessnumber);}if (JUDGE==0)    {        warning();        goto beg;    }    ed:return 0;}

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