LightOj 1214 Large Division（大数除法）

Large Division

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

解题思路：

模拟除法运算即可。

AC代码：

#include <bits/stdc++.h>using namespace std;typedef long long ll;int main(){    int T,t = 1;    scanf("%d",&T);    while(T--){        char num[300];        ll div;        scanf("%s%lld",num,&div);        printf("Case %d: ",t++);        if(div == 0){            printf("not divisible\n");            continue;        }        div = div>=0?div:-div;        if(num[0] == '-')            num[0] = '0';        int len = strlen(num);        ll ans = 0;        for(int i = 0; i < len; i++)            ans = ((num[i]-'0')+ans*10) % div;        if(ans == 0)            printf("divisible\n");        else            printf("not divisible\n");    }    return 0;}