LightOJ1214 Large Division 基础数论+同余定理

Posted 2020-07-03 ( m Lweleth)

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。

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <math.h>
 6 #define ll long long
 7 using namespace  std;
 8 
 9 char a[300];
10 int main()
11 {
12     int T, x, s;
13     ll t, b;
14     scanf("%d", &T);
15         for(int i=1; i<=T; i++)
16         {
17             scanf("%s %lld", a, &b);
18             x=strlen(a);
19             if(a[0]==‘-‘)//注意负数变正 
20             {
21                 s=2;
22                 t=a[1]-‘0‘;
23             }
24             else
25             {
26                 s=1;
27                 t=a[0]-‘0‘;
28             }
29             t=t%abs(b);
30             for(int j=s; j<x; j++)
31             {
32                 t=(t*10+a[j]-‘0‘)%abs(b); //同余定理的应用 
33             }
34 
35 
36             if(t==0)
37             {
38                 printf("Case %d: divisible\n", i);
39             }
40             else
41                 printf("Case %d: not divisible\n", i);
42 
43         }
44 
45 }