CODEWITHSUNDEEP
javaequation-solving
happy face
happy face

Reputation: 43

can someone explain the steps to compute this equation? Java

Write a program that computes the following equation. 100/1+99/2+98/3+97/4+96/5...3/98+2/99+1/100

I am not asking for a solution. Yes this is a homework problem, but I am not here to copy paste the answers. I asked my professor to explain the problem or how should I approach this problem? She said "I can't tell you anything." 

public static void main(String[] args){
    int i;

    for(i = 100; i >= 1; i--)
    {
        int result = i/j;
        j = j+1;
        System.out.println(result);
    }
}

Upvotes: 3

Views: 2610

Answers (4)

Rustam
Rustam

Reputation: 6515

if you want the result in the same format then do :

int j = 100;
double sum=0;
for (int i = 1; i <= 100; i++) {
sum += ((double) j / i);   // typecast as least one of i or j to double.
        System.out.print(j + "/" + i+"+");
  j--;
}

 // here print the sum

Upvotes: 0

Ed I
Ed I

Reputation: 7358

What you have is a series.

There is more than one way to define a series, but all things being the same it's more intuitive to have the index of a series increase rather than decrease.

In this case, you could use i from 0 to 99.

enter image description here

Which in java can be:

double sum = 0;
for (int i = 0; i < 100; i++) {
    sum += (100 - i) / (double) (1 + i);
}

Upvotes: 0

user3437460
user3437460

Reputation: 17454

You can try to observe a "trend" or "pattern" when solving questions of this type.

Given: 100/1+99/2+98/3+97/4+96/5...3/98+2/99+1/100

We derived: Numerator/Denominator, let's call it n divide by d (n/d)

Pattern Observed:

  • n - 1 after every loop
  • d + 1 after every loop

So, if you have 100 numbers, you need to loop 100 times. Thus using a for-loop which loops 100 times will seemed appropriate:

for(int n=0; n<100; n++) //A loop which loops 100 times from 0 - 100

To let n start with 100, we change the loop a little to let n start from 100 instead of 0:

for(int n=100; n>0; n--) //A loop which loops 100 times from 100 - 0

You settled n, now d needs to start from 1.

int d = 1; //declare outside the loop

Putting everything together, you get:

int d = 1;
double result = 0.0;
for (int n=100; n>0; x--)
{
    result += (double)n/d; //Cast either n or d to double, to prevent loss of precision
    d ++; //add 1 to d after every loop
}

Upvotes: 5

Alvin Bunk
Alvin Bunk

Reputation: 7764

You are on the right track. You need to loop like you've done, but then you need to SUM up all the results. In your example you can try:

result = result + i/j;

or

result += i/j;

Note that the declaration of result needs to be outside the loop otherwise you are always initializing it.
Also think about the division (hint), you are dividing integers...

Upvotes: 2

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