LightOJ Large Devision (大整數取模）

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 by b or not. We know that an integerais divisible by an integer b if and only if there exists an integercsuch 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) 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 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

思路：

         *  同餘：

                        (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;

         *大整數取模：

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

          *本題注意數據類型

代碼：

#include<cstdio>  
#include<cstring>    
typedef long long LL;
     
char s[301];//字符串儲存大整數
  
int main()  
{  
    int t;  
    scanf("%d", &t);    
   	for(int i=1; i<=t; i++)  
    {
	    LL ans=0;//注意數據類型
	    LL d; 
        scanf("%s%lld", s, &d);  
        int len=strlen(s);    
        for(int j=0; j<len; j++)  //可簡單理解爲模擬算術求餘過程
        {  
            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;  
}