题目:
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
第一种方法java水过(java自学过几天,看看博客,写一写)
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t;
BigInteger a,b,c = new BigInteger("0");
t = scan.nextInt();
for(int i=1;i<=t;i++)
{
a = scan.nextBigInteger();
b = scan.nextBigInteger();
if(a.remainder(b).equals(c))//取余,并把取余的结果与c(0)对比,如果结果是0,就可以
{
System.out.println("Case "+i+": divisible");
}
else{
System.out.println("Case "+i+": not divisible");
}
}
}
}
第二种c++,也是挺简单的,不断的进行取余;
#include<stdio.h>
#include<string.h>
int main()
{
int T,b,l;
char s[220];
int i,j;
long long sum;
scanf("%d",&T);
for(i=1; i<=T; i++) {
scanf("%s %d",s,&b);
l=strlen(s);
for(j=sum=0; j<l; j++) {
if(s[j]!='-') {
sum=(sum*10+s[j]-'0')%b;
}
}
if(!sum)
printf("Case %d: divisible\n",i);
else
printf("Case %d: not divisible\n",i);
}
return 0;
}