1.作业头
| 问题 | 答案 |
|---|---|
| 这个作业属于那个课程 | https://edu.cnblogs.com/campus/zswxy/CST2019-2 |
| 这个作业要求在哪里 | https://edu.cnblogs.com/campus/zswxy/CST2019-2/homework/10279 |
| 这个作业的目标 | 给上次作业的随机口算题提供答案 |
| 作业正文 | https://www.cnblogs.com/508775262-qq/p/12321566.html |
| 参考文献 | cpp6 百度 |
2.设计思路和遇到的问题
整体框架并无太大改动，因为考虑到是小学的问题，为了防止出现负数的情况，对一些特殊情况做了说明，并限制了一些数的范围。
3.程序结果截图
4.程序代码
#include
#include
#include
int menu()
{
printf("操作列表:\n 1)一年级 2)二年级 3)三年级\n 4)帮助 5)退出程序\n");
printf("请输入操作:> ");
return 0;
}
int help(){
printf("帮助信息\n您需要输入命令代号来进行操作, 且\n一年级题目为不超过十位的加减法;\n二年级题目为不超过百位的乘除法;\n三年级题目为不超过百位的加减乘除混合题目.\n\n");
return 0;
}
int error()
{
printf("\n\n");
printf("Error!!!\n错误操作指令, 请重新输入\n\n\n");
return 0;
}
void operation_1(){
srand(time(NULL));
int sum1,sum2,i,n;
char op[2]={'+','-'};
printf("现在是一年级题目：\n请输入生成题目的个数：");
scanf("%d",&n);
for(i=n;i>0;i--){
sum1=rand()%11;
sum2=rand()%11;
switch(op[rand()%2])
{
case '+': printf("%d + %d = %d\n",sum1,sum2,sum1+sum2); break;
case '-': if(sum10;i--){
sum1=rand()%101;
sum2=rand()%101;
c=op[rand()%2];
switch(op[rand()%2])
{
case'*':if(sum1==0||sum2==0)
printf("0 \n");
else
printf("%d * %d = %d\n",sum1,sum2,sum1*sum2);break;
case'/':if(c=='/'&&sum2==0)
sum2=rand()%100+1;
else if(sum1==0)
printf("0 \n");
else printf("%d / %d = %g\n",sum1,sum2,(double)sum1/sum2);
}
}
printf("\n\n");
}
void operation_3(){
srand(time(NULL));
int sum1,sum2,sum3,i,n;
char op[4]={'*','/','+','-'},op1,op2;
printf("现在是三年级题目：\n请输入生成题目的个数：");
scanf("%d",&n);
for(i=n;i>0;i--){
sum1=rand()%101;
op1=op[rand()%4];
sum2=rand()%11;
if(op1=='/'&&sum2==0)
sum2=rand()%10+1;
op2=op[rand()%4];
sum3=rand()%11;
if(op2=='/'&&sum3==0)
sum3=rand()%10+1;
switch(rand()%12)
{
case 1: printf("%d + %d - %d = %d\n",sum1,sum2,sum3,sum1+sum2-sum3); break;
case 2: printf("%d + %d * %d = %d\n",sum1,sum2,sum3,sum1+sum2*sum3); break;
case 3: printf("%d + %d / %d = %g\n",sum1,sum2,sum3,(double)sum1+sum2/sum3); break;
case 4: printf("%d - %d + %d = %d\n",sum1,sum2,sum3,sum1-sum2+sum3); break;
case 5: printf("%d - %d * %d = %d\n",sum1,sum2,sum3,sum1-sum2*sum3); break;
case 6: printf("%d - %d / %d = %g\n",sum1,sum2,sum3,(double)sum1-sum2/sum3); break;
case 7: printf("%d * %d + %d = %d\n",sum1,sum2,sum3,sum1*sum2+sum3); break;
case 8: printf("%d * %d - %d = %d\n",sum1,sum2,sum3,sum1*sum2-sum3); break;
case 9: printf("%d * %d / %d = %g\n",sum1,sum2,sum3,(double)sum1*sum2/sum3);break;
case 10: printf("%d / %d + %d = %g\n",sum1,sum2,sum3,(double)sum1/sum2+sum3); break;
case 11: printf("%d / %d - %d = %g\n",sum1,sum2,sum3,(double)sum1/sum2-sum3); break;
case 12: printf("%d / %d * %d = %g\n",sum1,sum2,sum3,(double)sum1/sum2*sum3); break;
}
}
printf("\n\n");
}
int main()
{
int Q;
printf("========== 口算生成器 ==========\n");
printf("欢迎使用口算生成器 :\n \n \n帮助信息\n您需要输入命令代号来进行操作, 且\n");
printf("一年级题目为不超过十位的加减法;\n二年级题目为不超过百位的乘除法;\n三年级题目为不超过百位的加减乘除混合题目.\n\n\n");
do{
menu();
scanf("%d",&Q);
printf("< 执行操作 >\n\n");
switch (Q) {
case 1:operation_1();break;
case 2:operation_2();break;
case 3:operation_3();break;
case 4:help();break;
case 5:printf("程序结束,欢迎下次使用任意键结束……\n");return 0;
default:error();break;
}
}while(Q!=5);
return 0;
} 5. Gitee上传截图与链接
链接：https://gitee.com/liu_qing_yu0103/lqy20199237