LightOJ 1214 - Large Division （大数求余）

Given two integers,a and b,you should check whethera is divisible by b or not. We know that an integera is divisible by an integer b if and only if there exists an integerc such that a = b * c.

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) andb (|b| > 0,b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.

For each case,print the case number first. Then print 'divisible' ifa is divisible by b. Otherwise print 'not divisible'.

6

101 101

0 67

-101 101

7678123668327637674887634 101

11010000000000000000 256

-202202202202000202202202 -101

Case 1: divisible

Case 2: divisible

Case 3: divisible

Case 4: not divisible

Case 5: divisible

Case 6: divisible

题意：问b是否能整除a

思路：大数求余，注意b会爆int

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 101000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,sizeof(x))
#define PI acos(-1)
using namespace std;
char s[MAXN];
int main()
{
	int n,i,t;
	LL b;
	int cas=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%lld",s,&b);
		if(b<0)
		b=-b;
		int len=strlen(s);
		LL num=0;
		for(i=0;i<len;i++)
		{
			if(s[i]=='-')
			continue;
			num=(num*10+s[i]-'0')%b;
		}
		printf("Case %d: ",++cas);
		if(num==0)
		printf("divisiblen");
		else
		printf("not divisiblen");
	}
	return 0;
}