LightOJ 1078 Integer Divisibility （同余定理）

LightOJ - 1078 

题目：

If an integer is not divisible by 2 or 5, some multiple of that number in decimal notation is a sequence of only a digit. Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.

For example you have to find a multiple of 3 which contains only 1's. Then the result is 3 because is 111 (3-digit) divisible by 3. Similarly if you are finding some multiple of 7 which contains only 3's then, the result is 6, because 333333 is divisible by 7.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case will contain two integers n (0 < n ≤ 106 and n will not be divisible by 2 or 5) and the allowable digit (1 ≤ digit ≤ 9).

Output

For each case, print the case number and the number of digits of such multiple. If several solutions are there; report the minimum one.

Sample Input

3

3 1

7 3

9901 1

Sample Output

Case 1: 3

Case 2: 6

Case 3: 12

题意：给出 n和d，让判断多少位的d可以整除n（每一位的数值都为d，如果当前的位数不能整除n则再加上一位）

分析：

可怜的java大数超时

正解 同余定理

(A + B) mod M = ( A mod M + B mod M ) mod M
(A * B) mod M = ((A mod M) *( B mod M)) mod M  

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e5+5;
int a[N];
 
int main(){
    int T;
    scanf("%d",&T);
    int cas=0;
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        int ans=m;
        for(int i=1;;i++)
        {
            if(ans%n==0)
            {
                ans=i;
                break;
            }
            else
            {
                ans=(ans*10+m)%n;
            }
        }
        printf("Case %d: %d\n",++cas,ans);
       
    }
}

import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main {
    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner (new BufferedInputStream(System.in));
        //Scanner in = new Scanner(new BufferedInputStream(System.in));
        Reader.init(System.in);
        int T=in.nextInt();
        int cas=0;
        while(T-->0)
        {
            cas++;
         BigInteger n=in.nextBigInteger();  
         BigInteger m=in.nextBigInteger();
      
            BigInteger t=m;
            int i=1;
            for(i=1;;i++)
            {
                
                
                BigInteger t1=t;//System.out.println(t1.mod(n));
                if(t.mod(n).equals(new BigInteger("0")))
                {
                    break;
                }
                else
                {
                    t=t.multiply(new BigInteger("10")).add(m);
                }
            }
            System.out.println("Case "+cas+": "+i+"\n");
     

    
        }
        
    }

    
}
class Reader 
{ 
    static BufferedReader reader; 
    static StringTokenizer tokenizer; 
 
    static void init(InputStream input) 
    { 
        reader = new BufferedReader(new InputStreamReader(input)); 
        tokenizer = new StringTokenizer(""); 
    } 
 
    static String next() throws IOException 
    { 
        while(!tokenizer.hasMoreTokens()) 
        { 
            String str=reader.readLine();
            //System.out.println(str);
            tokenizer = new StringTokenizer(str); 
        } 
        return tokenizer.nextToken(); 
    } 
 
    static int nextInt() throws  IOException 
    { 
        return Integer.parseInt(next()); 
    } 
 
    static double nextDouble() throws IOException 
    { 
        return Double.parseDouble(next()); 
 
    } 
}