【light-oj】-大数整除（字符串）

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

#include <stdio.h>
#include <string.h>
int main()
{	
	int n;
	char a[300];
	int m;
	long long y;
	int l;
	int t=1;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%s %d",a,&m);
		l = strlen(a);
		if(m < 0)
			m = -m;				 			//m作为除数不可以为负  
		printf("Case %d: ",t++);
		if(a[0]=='-')					//m分正负单独处理
		{
			y=(a[1]-'0')%m; 		
			for(int i=2;i<l;i++)
			{
				y=(y*10+(a[i]-'0'))%m;
			}
			if(y==0)
				printf("divisiblen");
			else
				printf("not divisiblen");
		}
		else
		{
			y=(a[0]-'0')%m;
			for(int i=1;i<l;i++)
			{
				y=(y*10+(a[i]-'0'))%m;
			}
			if(y==0)
				printf("divisiblen");
			else
				printf("not divisiblen");
		}
	}
	return 0;
}

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