light oj 1214 - Large Division 大数除法

1214 - Large Division

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

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define N 1000010

using namespace std;
char ar[N];
int nu[N];
long long m;
void option()
{
long long x, y;
int j;
int i = 0;
int len = strlen(ar);
if(ar[0] == '-')
{
len--;
i = 1;///判断是否为负数，i为ar数组的起始下标。
}
for( j = 0; ar[j]; j++, i++)
nu[j] = ar[i] - '0';///将字符串转换为数字。

if(m < 0)
m = -m;

for(j = 0, y = 0; j < len; )
{
x = y;

while(x <= m && j < len)
{
x = x * 10 + nu[j];
j++;
}

y = x % m;
}

if(y)
printf("not divisible\n");
else
printf("divisible\n");
}
int main(void)
{
int T, cas;
scanf("%d", &T);
cas = 0;
while(T--)
{
cas++;
scanf("%s%lld", ar, &m);
printf("Case %d: ", cas);
option();
}
return 0;
}