【Lightoj】1214 - 能否整除（同余定理）

Time Limit:1000MS
Memory Limit:32768KB 64bit IO Format:%lld & %llu

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Status
Practice
LightOJ 1214

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

简单同余定理。

代码如下：

#include <stdio.h>
#include <string.h>
int main()
{
	int u;
	char t[222];
	int a[222];
	long int b;
	int l;		//数字总位数 
	long int y;		//余数 
	long int ty;		//临时存放余数 
	int i,j;
	scanf ("%d",&u);
	getchar();
	for (int p=1;p<=u;p++)
	{
		
		memset (a,0,sizeof (a));
		memset (t,'0',sizeof (t));
		scanf ("%s",t);
		scanf ("%ld",&b);
		printf ("Case %d: ",p);
		if (b<0)	b=-b;
		l=strlen(t);
		if (t[0]=='-')
		{
			l--;
			for (i=1,j=l;i<=l;i++,j--)
			{
				a[i]=t[j]-48;
			}
		}
		else
		{
			for (i=1,j=l-1;i<=l;i++,j--)
			{
				a[i]=t[j]-48;
			}
		}
		y=a[1]%b;
		for (i=2;i<=l;i++)
		{
			if (a[i]==0)	continue;
			ty=a[i]%b;
			for (j=2;j<=i;j++)
			{
				ty=(ty*10)%b;
			}
			y=(y+ty)%b;
		}
		if (y==0)
		{
			printf ("divisible\n");
		}
		else
		{
			printf ("not divisible\n");
		}
	}
	return 0;
}