對二柱子的專案完成更新
阿新 • • 發佈：2018-11-07
題目要求：1，實現線上答題；
2，答題結束後，可以判斷對錯；
3，並將錯題的結果儲存起來。
實驗過程：
#include<iostream.h>#include<stdlib.h>
#include<time.h>
void boom()
{
float x,y;
int z,t;
int first_num,second_num,third_num,fourth_num;
srand(time(NULL));
int n;
float k;
float d;
float daan[100];
int zq=0;
int cw=0;
cout<<"需要列出多少到數學題?"<<endl;
cin>>n;
cout<<"********************************"<<endl;
for(int i=0;i<n;i++)
{
z=rand()%4;
first_num=rand()%10;
second_num=rand()%10;
third_num=rand()%10;
fourth_num=rand()%10;
if (second_num==1)
{
x=(float)first_num;
}
else if(first_num>second_num)
{
t=first_num;
first_num=second_num;
second_num=t;
}
x=(float)first_num/second_num;
if (fourth_num==1)
{
y=(float)third_num;
}
else if(third_num>fourth_num)
{
t=third_num;
third_num=fourth_num;
fourth_num=t;
}
y=(float)third_num/fourth_num;
z=rand()%4;
switch(z)
{
case 0:
cout<<i+1<<"、 "<<first_num<<"/"<<second_num<<"+"<<third_num<<"/"<<fourth_num<<"="<<" ";
daan[i+1]=x+y;
break;
case 1:
cout<<i+1<<"、 "<<first_num<<"/"<<second_num<<"-"<<third_num<<"/"<<fourth_num<<"="<<" ";
daan[i+1]=x-y;
break; case 2:
cout<<i+1<<"、 "<<first_num<<"/"<<second_num<<"*"<<third_num<<"/"<<fourth_num<<"="<<" ";
daan[i+1]=x*y;
break;
case 3:
if(y!=0)
{
cout<<i+1<<"、 "<<first_num<<"/"<<second_num<<"÷"<<third_num<<"/"<<fourth_num<<"="<<" ";
}
else
{
i=i-1;
}
daan[i+1]=(float)x/y;
break;
}
if((i+1)%3==0)
{
for(int c=0;c<=1;c++)
{
cout<<endl;
}
}
if((i+1)==n)
{
cout<<endl;
}
}
for(i=1;i<=n;i++)
{
cout<<"請作答:"<<endl;
cout<<i<<"、";
cin>>d;
if(d==daan[i])
{
cout<<"恭喜您，回答正確！"<<endl;
cout<<endl;
zq=zq+1;
}
else
{
cout<<"回答錯誤，正確答案為："<<" "<<daan[i]<<endl;
cout<<endl;
cw=cw+1;
}
}
k=(float)zq/n;
cout<<"回答正確的數量為："<<" "<<zq<<"個"<<endl;
cout<<"回答錯誤的數量為："<<" "<<cw<<"個"<<endl;
cout<<"正確率為："<<" "<<k<<endl;
}
int main()
{
boom();
int c;
cout<<"要繼續答題嗎？（0：繼續 1：退出）"<<endl;
cin>>c;
if(c==0)
{
cout<<endl;
main();
}
else
{
cout<<endl;
return 0;
}
return 0;
}
實驗結果： 
