4. Write a program in java to enter any three subject marks and find average and display
grade rccording to
Grade
A
B
More than equals to 90
More than equals to "0 and less than 90
More than equals to 50 and less than 10
Othervise
с
F
Answers
Answer:
think f thi a suggestion to be honest with my name of a nice evening I am looking to buy it would have a lot to learn how you feel about a month and he will have no problem aahe to you have received the following user to you and I will get a better way of life in this thread
assuming that u want a solution that is faster computationally,
import java.io.*;
import java.util.*;
//don't concentrate on the FastReader class. move to the main method.
class Main{
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}
catch (IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int readInt(){
return Integer.parseInt(next());
}
}
public static void main(String[] args) {
FastReader scanner = new FastReader();
int num1 = Math.abs(scanner.readInt());int num2 = Math.abs(scanner.readInt());
int num3 = Math.abs(scanner.readInt());
double average = (double) (num1 + num2 + num3)/3;
if(num1<= 100&num2<=100;num3<=100){
if(average >= 90){
System.out.println("A");
}
else if(average < 90 & average >= 50){ /*more than equal to 0 not required
since abs value taken*/
//more than equal to 50 and less than 10 always returns false
System.out.println("B");
}
else {
if ((average >= 10 & average < 50)) {
System.out.println("C");
} else {
System.out.println("F");
}
}
}
else {
System.err.println("Invalid input");
}
}
}
if u find this answer helpful, please mark this as the brainliest.
if u find any of the lines confusing reach me out in the comment section