《软件工程》个人项目二：增加功能后的四则运算“软件”

Posted 2021-01-11 Surveyor

要求：

1.支持真分数运算；

2.一次出的题目避免重复；

3.可定制出题数目；

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define MAX 100
void simple(int &m,int &n) //约去分子分母公约数
{
int p;
p=(m+n-abs(m-n))/2;
for(int i=2;i<=p;i++)
if(m%i==0&&n%i==0)
{
m/=i;
n/=i;
i=2;
}
}

void show(int &x,int &y) //输出整数或分数
{
if(x%y==0)
printf("%d",x/y);
else
printf("%d/%d",x,y);
}

static int number=0;
void judge(int p,int q,int f,int g) //判断输入结果的对错
{
if(p==f&&q==g)
{
printf("回答正确!\\n");
number+=1;
}
else
printf("回答错误!\\n");
}

int main()
{
FILE *fp;
//文件的打开
if((fp=fopen("mytiku.txt","w"))==NULL)
{
printf("Cannot open this file!\\n");
exit(0);
}
for(int i=0;i<MAX;i++)
{
int a,b,c,d,e,f,g,p,q,x,y;
int *A=(int *)malloc(MAX*sizeof(int)),*B=(int *)malloc(MAX*sizeof(int));
int *C=(int *)malloc(MAX*sizeof(int)),*D=(int *)malloc(MAX*sizeof(int));
a=rand()%MAX+1;
b=rand()%MAX+1;
c=rand()%MAX+1;
d=rand()%MAX+1;
e=rand()%4+1;
simple(a,c);
simple(b,d);
A[i]=a;
B[i]=b;
C[i]=c;
D[i]=d;

//保证没有重复的题目
for(int j=0;j<i;j++)
if((A[i]==A[j]&&B[i]==B[j]&&C[i]==C[j]&&D[i]==D[j])||(A[i]==B[j]&&B[i]==A[j])&&C[i]==D[j]&&D[i]==C[j])
{
a=rand()%MAX+1;
b=rand()%MAX+1;
c=rand()%MAX+1;
d=rand()%MAX+1;
simple(a,c);
simple(b,d);
j=0;
}

//显示产生的算式
show(a,c);
switch(e)
{
case 1: printf("+");break;
case 2: printf("-");break;
case 3: printf("*");break;
case 4: printf("/");break;
}
show(b,d);
printf("=");

//对结果进行判断
switch(e)
{
case 1: f=a*d+b*c;
g=c*d;
simple(f,g);
if(g!=1)
{
scanf("%d/%d",&p,&q);
simple(p,q);
judge(p,q,f,g);
}
else
{
scanf("%d",&p);
judge(p,1,f,1);
}
fprintf(fp,"%d/%d+%d/%d=%d/%d\\n",a,c,b,d,f,g);break;
//减法可能产生负数，要对其进行判断
case 2: f=abs(a*d-b*c);
g=c*d;
x=a*d;
y=b*c;
simple(f,g);
if(x>=y)
{
if(g!=1)
{
scanf("%d/%d",&p,&q);
simple(p,q);
judge(p,q,f,g);
}
else
{
scanf("-%d",&p);
judge(p,1,f,1);
}
fprintf(fp,"%d/%d-%d/%d=%d/%d\\n",a,c,b,d,f,g);
}
else
{
if(g!=1)
{
scanf("-%d/%d",&p,&q);
simple(p,q);
judge(p,q,f,g);
}
else
{
scanf("%d",&p);
judge(p,1,f,1);
}
fprintf(fp,"%d/%d-%d/%d=-%d/%d\\n",a,c,b,d,f,g);
}break;
case 3: f=a*b;
g=c*d;
simple(f,g);
if(g!=1)
{
scanf("%d/%d",&p,&q);
simple(p,q);
judge(p,q,f,g);
}
else
{
scanf("%d",&p);
judge(p,1,f,1);
}
fprintf(fp,"%d/%d*%d/%d=%d/%d\\n",a,c,b,d,f,g);break;
case 4: f=a*d;
g=b*c;
simple(f,g);
if(g!=1)
{
scanf("%d/%d",&p,&q);
simple(p,q);
judge(p,q,f,g);
}
else
{
scanf("%d",&p);
judge(p,1,f,1);
}
fprintf(fp,"%d/%d/%d/%d=%d/%d\\n",a,c,b,d,f,g);break;
}
}
fprintf(fp,"做对题目的数目为%d\\n",number);
fclose(fp);
return 0;
}

结果：