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

 

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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 longusing 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;}