LightOJ Large Devision (大整数取模）

1214 - Large Division

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Time Limit: 1 second(s)Memory Limit: 32 MB

Given two integers, a and b, you should checkwhethera is divisible byb or not. We know that an integerais divisible by an integerb if and only if there exists an integercsuch thata = 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) andb (|b| >0, b fits into a 32 bit signed integer). Numbers will not contain leadingzeroes.

Output

For each case, print the case number first. Then print 'divisible'ifa is divisible byb. 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

思路：

         *  同余：

                        (1). (a + b )% n = (a%n +b%n ) % n

                         (2) (a - b ) % n =(a%n -b%n +n)%n

                         (3)  ab%n  =  (a%n)(b%n)%n;

         *大整数取模：

                   如 1234=((1*10+2)*10+3)*10+4) 按照1.3公式每步取模。

          *本题注意数据类型

代码：

#include<cstdio>  #include<cstring>    typedef long long LL;     char s[301];//字符串储存大整数  int main()  {      int t;      scanf("%d", &t);       for(int i=1; i<=t; i++)      {    LL ans=0;//注意数据类型    LL d;         scanf("%s%lld", s, &d);          int len=strlen(s);            for(int j=0; j<len; j++)  //可简单理解为模拟算术求余过程        {              if(s[j]=='-')  continue;              ans=(ans*10+s[j]-'0')%d;          }          printf("Case %d: ",i);          if(ans) printf("not divisible\n");          else     printf("divisible\n");      }      return 0;  }