标签:lightoj 1214 大数相除
Description
Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c.
Input
Input starts with an integer T (≤ 525), denoting the number of test cases.
Each case starts with a line containing two integers a (-10200 ≤ a ≤ 10200) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.
Output
For each case, print the case number first. Then print ‘divisible‘ if a is divisible by b. Otherwise print ‘not divisible‘.
Sample Input
6
101 101
0 67
-101 101
7678123668327637674887634 101
11010000000000000000 256
-202202202202000202202202 -101
Sample Output
Case 1: divisible
Case 2: divisible
Case 3: divisible
Case 4: not divisible
Case 5: divisible
Case 6: divisible
直接上Java
import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
BigInteger a, b;
int t;
t = in.nextInt();
for (int i = 1; i <= t; i++) {
a = in.nextBigInteger();
b = in.nextBigInteger();
if(b.compareTo(BigInteger.ZERO)<0) b=b.multiply(BigInteger.valueOf(-1));
BigInteger c = a.mod(b);
if (c == BigInteger.ZERO)
System.out.println("Case " + i + ": " + "divisible");
else
System.out.println("Case " + i + ": " + "not divisible");
}
}
}标签:lightoj 1214 大数相除
原文地址:http://blog.csdn.net/acm_baihuzi/article/details/46539999