LightOJ1214 Large Division —— 大数求模

题目链接：https://vjudge.net/problem/LightOJ-1214

1214 - Large Division

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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

题解：

单纯的大数求模。从高位处理到低位。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int mod = 1e9+;
 const int MAXM = 1e5+;
 const int MAXN = 5e5+;
 
 char a[];
 int b;
 
 int main()
 {
     int T, kase = ;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%s%d", a, &b);
         int len = strlen(a);
         b = abs(b);
         LL s = ;
         for(int i = (a[]=='-'); i<len; i++)
         {
             s *= , s += a[i] - '';
             s %= b;
         }
         printf("Case %d: %s\n", ++kase, s?"not divisible":"divisible");
     }
 }