所以这个程序可以根据你输入的百分比告诉你你的成绩，我基本上只使用if和else语句对它进行编码。该程序一直工作到59％，然后当我输入任何超过该百分比的东西时，它不起作用。因为在程序中只是不会告诉我59％之后的成绩。没有任何东西可以帮助!!!
P.S。我知道可能有一种更简单的方法可以让我编程，但我想练习if和else语句......
//
//exercise 1.
//you will enter in your percent and it will anounce your grade.
//create a program so that it will notify the user of their letter grade
//0-59 F 60-69 D 70-79 C 80-89 B 90-100 A
//a=user input
//For whatever reason the program seems to only work up to 59% and after that it doesn't work.
int a;
#include <iostream>
using namespace std;
int main()
{
cout<< "Enter in the percent of your grade and \n I will tell you your grade"<<endl;
cin>>a;
if(a==100)
{
cout<<"you scored a perfect A";
}
else
if(a<=59)
{
if(a<0)
{
cout<<"your really stupid";
}
else
cout<<"you failed";
}
else
if(a>=60)
{
if(a<=69)
{
cout<<"You got a D";
}
}
else
if(a>=70)
{
if(a<=79)
{
cout<<"you got a C";
}
}
else
if(a>=80)
{
if(a<=89)
{
cout<<"you got a B";
}
}
else
if(a>=90)
{
cout<<"you got an A";
}
}
答案 0 :(得分：1)
问题是你正在检查一个＆gt; = 60 ......它是不是（假设a = 75）。但是，它没有被任何其他条件语句捕获。通过以下代码中的评论更好地解释。
#include <iostream>
using namespace std;
int a;
int main()
{
cout<< "Enter in the percent of your grade and \n I will tell you your grade"<<endl;
cin>>a;
if(a==100)
{
cout<<"you scored a perfect A";
}
else
if(a<=59)
{
if(a<0)
{
cout<<"your really stupid";
}
else
cout<<"you failed";
}
else
if(a>=60) // 75 >= 60
{
if(a<=69) // But 75 is > 69
{
cout<<"You got a D";
}
}
// ONLY REACHES THIS POINT IF a < 60
else
if(a>=70)
{
if(a<=79)
{
cout<<"you got a C";
}
}
else
if(a>=80)
{
if(a<=89)
{
cout<<"you got a B";
}
}
else
if(a>=90)
{
cout<<"you got an A";
}
}
这可能是一种更好的方法：
int main()
{
cout<< "Enter in the percent of your grade and \n I will tell you your grade"<<endl;
cin>>a;
if(a==100)
{
cout<<"you scored a perfect A";
}
else
{
if(a<=59)
{
if(a<0)
{
cout<<"your really stupid";
}
else
cout<<"you failed";
}
else
{
if (a >= 60 && a <= 69)
{
cout << "D";
}
else if (a >= 70 && a <= 79)
{
cout << "C";
}
else if (a >= 80 && a <= 89)
{
cout << "B";
}
else
{
cout << "A";
}
}
}
}
答案 1 :(得分：0)
我认为你在第一个之后缺少括号。
试试这个：
//
//exercise 1.
//you will enter in your percent and it will anounce your grade.
//create a program so that it will notify the user of their letter grade
//0-59 F 60-69 D 70-79 C 80-89 B 90-100 A
//a=user input
//For whatever reason the program seems to only work up to 59% and after that it doesn't work.
int a;
#include <iostream>
using namespace std;
int main()
{
cout<< "Enter in the percent of your grade and \n I will tell you your grade"<<endl;
cin>>a;
if(a==100)
{
cout<<"you scored a perfect A";
}
else
{
if(a<=59)
{
if(a<0)
{
cout<<"your really stupid";
}
else
cout<<"you failed";
}
else
if(a>=60)
{
if(a<=69)
{
cout<<"You got a D";
}
}
else
if(a>=70)
{
if(a<=79)
{
cout<<"you got a C";
}
}
else
if(a>=80)
{
if(a<=89)
{
cout<<"you got a B";
}
}
else
if(a>=90)
{
cout<<"you got an A";
}
}
} // end main
答案 2 :(得分：0)
在您的代码中，一旦到达if(a>=60)，如果a的值不小于70，则不输出任何内容，并且代码退出条件，因为它找到了真实的情况。至于为什么你不能让它在59以上做任何事情，我不知道，因为它看起来应该有效。但是，通过将内部if块与其父项组合，您可以获取代码以检查a是否在10个百分点的组内，如果它不在该组中，则转到下一个。以下是如何执行此操作的示例。
//
//exercise 1.
//you will enter in your percent and it will anounce your grade.
//create a program so that it will notify the user of their letter grade
//0-59 F 60-69 D 70-79 C 80-89 B 90-100 A
//a=user input
//For whatever reason the program seems to only work up to 59% and after that it doesn't work.
int a;
#include <iostream>
using namespace std;
int main()
{
cout<< "Enter in the percent of your grade and \n I will tell you your grade"<<endl;
cin>>a;
if(a==100)
{
cout<<"you scored a perfect A";
}
else if(a<=59)
{
if(a<0)
{
cout<<"your really stupid";
}
else
{
cout<<"you failed";
}
}
else if(a>=60 && a<=69)
{
cout<<"You got a D";
}
else if(a>=70 && a<=79)
{
cout<<"you got a C";
}
else if(a>=80 && a<=89)
{
cout<<"you got a B";
}
else if(a>=90)
{
cout<<"you got an A";
}
}