Large Division（大数取余）

Link：https://vjudge.net/contest/172540#problem/L

题目：

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

6
101 101
0 67
-101 101
7678123668327637674887634 101
11010000000000000000 256
-202202202202000202202202 -101

Sample Output

Case 1: divisible
Case 2: divisible
Case 3: divisible
Case 4: not divisible
Case 5: divisible
Case 6: divisible

思路：

同余定理：

    根据同余定理可得出：(a+b)%c=((a%c)+(b%c))%c

a%c举例：

对于一个大数求其%N 的值，例如求1234 可以写成 1*1000+2*100+3*10+4 =((1*10)+2)*10+3)*10+4 所以 1234%N=(((1*10)%N+2)*10%N+3)*10%N+4%N

程序关键点：

设一整型数据ans存储结果 ans=0；
ans=(ans*10+a[i])%N (a[i] 代表数据每个位上的数据)

Code：

#include
#include
int main() 
{
    int T,b,l;
    char s[220];
    int i,j;
    long long sum;
    scanf("%d",&T);
    for(i=1; i<=T; i++) {
        scanf("%s %d",s,&b);
        l=strlen(s);
        for(j=sum=0; jif(s[j]!='-') {
                sum=(sum*10+s[j]-'0')%b;
            }
        }
        if(!sum)
            printf("Case %d: divisible\n",i);
        else
            printf("Case %d: not divisible\n",i);
    }
    return 0;
}