수다닷컴

질문그대로...
나름 짰는데 너무 긴것 같아서 좀 짧게 할 수 없을까요...?
굵게 적힌 부분이 4번이나 있는데요 너무 복잡한거 같아서요..
함수를 이용해보려고 했는데 좀 어설픈실력인데다가 해봤는데 변수가 계속 문제라...
그리고 밑에 문제가 나와서 풀면 맞는개수가 몇개인지 표시해야되는데
어떻게해야할까요? 힌트좀..
질문 내용 :
#include stdio.h
#include time.h
#include stdlib.h
void main()
{
int q, w; //q는 문제 숫자. w는 문제 뭐 풀지.
int a1, a2, ans; // a1은 숫자1, a2는 숫자 2. ans는 답.
int b;
printf(how many want to question? : );
scanf_s(%d, &q);
printf(choose the question\n1)addition\n2)subtraction\n3)multiplication\n4)division\n);
scanf_s(%d, &w);
if(w4){
printf(wrong number...-_-;;;;\n);
return;
}
else{}
printf(\n\nwrite down answer.\n);
if(w==1){
for(b=1;bq+1;b++){
srand(time(null));
a1 = rand() % 100+1;
a2 = rand() % 100+1;

printf(%d + %d = , a1, a2);
scanf_s(%d, &ans);
if((a1+a2)==ans){printf(the answer is very good.\n);}
else{printf(the answer is very bad.\n);}
}
}
else if(w==2){
for(b=1;bq+1;b++){
srand(time(null));
a1 = rand() % 100+1;
a2 = rand() % 100+1;
printf(%d - %d = , a1, a2);
scanf_s(%d, &ans);
if((a1-a2)==ans){printf(the answer is very good.\n);}
&nbnbsp;else{printf(the answer is very bad.\n);}
}
}
else if(w==3){
for(b=1;bq+1;b++){
srand(time(null));
a1 = rand() % 10+1;
a2 = rand() % 10+1;
printf(%d * %d = , a1, a2);
scanf_s(%d, &ans);
if((a1*a2)==ans){printf(the answer is very good.\n);}
else{printf(the answer is very bad.\n);}
}
}
else if(w==4){
for(b=1;bq+1;b++){
srand(time(null));
a1 = rand() % 10+1;
a2 = rand() % 10+1;
printf(%d ÷ %d = , a1, a2);
scanf_s(%d, &ans);
if(((float)a1/a2)==ans){printf(the answer is very good.\n);}
else{printf(the answer is very bad.\n);}
}
}
}

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