lightoj 1214 - Large Division 【判大数整除 拆分字符串 同余】

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

 

PROBLEM SETTER: JANE ALAM JAN

题意：给你一个a (-10^100 <= a <= 10^100)，又给你一个b(int范围)，问你b能否整除a。

思路：用字符串表示a，然后依次拆分每一位，根据同余性质对b取余到最后一位，判断最后结果是否为0。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN 500000+10
#define MAXM 50000000
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
using namespace std;
int main()
{
int t, kcase = 1;
Ri(t);
W(t)
{
LL mod; char str[200];
Rs(str); Rl(mod);
int s = 0;
int len = strlen(str);
if(str[0] == '-')
s++;
LL ans = 0;
if(mod < 0)
mod = -mod;
for(int i = s; i < len; i++)
ans = (ans * 10 + str[i]-'0') % mod;
if(ans == 0)
printf("Case %d: divisible\n", kcase++);
else
printf("Case %d: not divisible\n", kcase++);
}
return 0;
}