LightOJ1214 - Large Division（大数取余）

题解：

同余定理，注意b可能爆int

代码

#include <cstdio>#include <queue>#include <cstring>#include <iostream>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <string>#include <set>#include <ctime>#include <cmath>#include <cctype>using namespace std;#define maxn 100000#define LL long longint cas=1,T;char s[maxn];int main(){    scanf("%d",&T);    while (T--)    {        LL b;        scanf("%s%lld",s,&b);        if (b<0);          b= -b;        int len = strlen(s);        LL num = 0;        for (int i = 0;i<len;i++)        {            if (s[i] == '-')                continue;            num = (num*10+s[i]-'0') % b;        }        printf("Case %d: ",cas++);        if (!num)            printf("divisible\n");        else            printf("not divisible\n");    }    //freopen("in","r",stdin);    //scanf("%d",&T);    //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);    return 0;}

题目

Description

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