LightOJ1214 Large Division —— 大数求模

题目链接：https://vjudge.net/problem/LightOJ-1214

1214 - Large Division

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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	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

题解：

单纯的大数求模。从高位处理到低位。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int MAXM = 1e5+10;
const int MAXN = 5e5+10;

char a[220];
int b;

int main()
{
    int T, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s%d", a, &b);
        int len = strlen(a);
        b = abs(b);
        LL s = 0;
        for(int i = (a[0]=='-'); i<len; i++)
        {
            s *= 10, s += a[i] - '0';
            s %= b;
        }
        printf("Case %d: %s
", ++kase, s?"not divisible":"divisible");
    }
}