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 long
int 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