LightOj 1214 Large Division（大数除法）

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本文介绍了一种处理大数除法的方法，并提供了一个具体的编程实现案例。通过模拟除法运算，判断大数a是否能被较小整数b整除，并给出相应结果。

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

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;
}