LightOJ1214 Large Division 基础数论+同余定理

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	Output for Sample Input
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：给出两个数a, b，问能否被b整除。

题解：基础数论。简单的同余定理应用，将a作为字串储存，相当于每x位(和b同位)模b一次，得到余数时相当于将这个区间改写成这个余数，移动区间继续运算。最终余数为零时代表可以被整除，非零则否。

补充：其实可以想成这样，三位数就是百进制，四位数就是千进制的同余定理。

 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 #include <iostream>
 #include <math.h>
 #define ll long long
 using namespace  std;

 char a[];
 int main()
 {
     int T, x, s;
     ll t, b;
     scanf("%d", &T);
         for(int i=; i<=T; i++)
         {
             scanf("%s %lld", a, &b);
             x=strlen(a);
             if(a[]=='-')//注意负数变正
             {
                 s=;
                 t=a[]-'';
             }
             else
             {
                 s=;
                 t=a[]-'';
             }
             t=t%abs(b);
             for(int j=s; j<x; j++)
             {
                 t=(t*+a[j]-'')%abs(b); //同余定理的应用
             }

             if(t==)
             {
                 printf("Case %d: divisible\n", i);
             }
             else
                 printf("Case %d: not divisible\n", i);

         }

 }