LightOj 1214 Large Division 【大数取模】

思路：一个大数对一个数取余，可以把大数看成各位数的权值与个位数乘积的和。比如1234=（ （1*10 + 2 ）* 10 +3 ）* 10+4，对这个数进行取余运算就是上面基本加和乘的应用。

代码实现：

int len = a.length();
int ans=0;
for( int i=0; i<len; i++){
	ans = ( ans * 10 + a[i] - '0' ) mod b;
}

举个LightOj的例子，代码实现如下：

#include<bits/stdc++.h>
using namespace std;
int main(){
	char s[100050];
	long long b,res=1;
	int t;
	cin>>t;
	while(t--){
		long long ans=0;
		cin>>s>>b;
		int l=strlen(s);
		for(int i=0;i<l;i++){
			if(s[i]=='-')
			    continue;
			ans=(ans*10+s[i]-'0')%b;
		}
		//printf("Case %d: ",res++);
		cout<<"Case "<<res++<<": ";
		if(ans)
		  cout<<"not divisible"<<endl;
		else
		  cout<<"divisible"<<endl;
	}
	return 0;
}

ps：我想带你一起去看海！

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