CODEWITHSUNDEEP
java
司徒残影
司徒残影

Reputation: 35

BigInteger Wrong Answer

I was calculating a simple combination using java, when I define the result as double, the answer seems right, only lost some precision, but when I use the BigInteger class, the answer looks just wrong at all, I don't see why it's like that, here is the code.

import java.math.BigInteger;
import java.util.Scanner;


public class Test {

public static void main(String[] args){

    Scanner in = new Scanner(System.in);
    System.out.println("Please input the lottery sum");
    int num = in.nextInt();
    System.out.println("Please input the available sum");
    int avail = in.nextInt();

    //double sum = 1;
    BigInteger sum = BigInteger.ONE;

    for (int i = avail - 1; i >= 1;i--){
        //sum = sum*(num - i) / i;
        sum = sum.multiply(BigInteger.valueOf(num - i)).divide(BigInteger.valueOf(i));
    }

    //sum = sum * num / avail;
    sum = sum.multiply(BigInteger.valueOf(num)).divide(BigInteger.valueOf(avail));
    System.out.println(sum);

    in.close();
}
}

Upvotes: 1

Views: 346

Answers (2)

steveman
steveman

Reputation: 139

I'm not sure what you're calculating, but I suspect that the integer division is what's causing the observed differences.

I suppose that you get the same result with BigInteger and a regular int for small examples.

For example: double: 3.0/2.0 = 1.5

(Big)Integer: 3/2 = 1

If you sum over such divisions multiple times, the discrepancy gets bigger and bigger.

Take a look at BigDecimal, that may be the solution you're after to avoid this error introduced by integer divisions.

Upvotes: 0

NPE
NPE

Reputation: 500257

The main difference between double and BigInteger is that one is using float-point maths, and other is using integer maths. This means that, when you using BigInteger, the result of every division gets truncated.

Upvotes: 3

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