【练习赛补题】1214

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

题意：让你找出一个大数能否被另一个数整除
思路：模拟除法运算过程，比如114除以3，我们先用11%3== 2，再用（2*10+4）%3==0，取余为0时，说明能够整除。还有同余定理的另一种写法，放在最后。由于第一个数范围很大，所以我们用字符串读入，第二个数的读入，我以为是用int，wrong了一次，去oj上的讨论版看数据，结果发现有一个数据不能过，就是下面给出这个，改为onglong才是正解，因为这个数据非常特殊。是2^31-1，刚好是题目边界数据，这时候我们*10就超int范围，虽然都是同余定理的应用，但是感觉这道题poj上1426好理解很多

今天早上最大的收获当然是light  oj~~~ ，因为赵老师在上面找的题才发现这个oj特别棒，讨论版的话是全英的，不过只要英语不是特别弱应该都能看懂的

1

10737418235 2147483647

方法1

#include<stdio.h>
#define N 2100
char s1[N];
long long  sum,d;

int main()
{
    int t,i,t2=0,flag;
    char *p;
    scanf("%d",&t);getchar();
    while(t --)
    {
        i = flag = 0;
        scanf("%s %lld",s1,&d);
        p = s1;
        if(*p =='-')
            *p ++;//puts(p);
        sum = p[i]-'0';
        while(p[i]!='')
        {
            if(sum %d == 0&&p[i+1]=='')
            {
                flag = 1;
                break;
            }
            sum = (sum%d)*10 + p[++i]-'0';
        }
        printf("Case %d: ",++t2);
        if(flag)
            printf("divisible
");
        else
            printf("not divisible
");
    }
    return 0;
}

方法2

#include<stdio.h>
#include<string.h>
#define N 2100
char s1[N];
long long  sum,d;

int main()
{
    int t,i,t2=0;
    char *p;
    scanf("%d",&t);getchar();
    while(t --)
    {
        i = 0;
        scanf("%s %lld",s1,&d);
        p = s1;
        if(*p =='-')
            *p ++;
        sum = 0;
        while(i < strlen(p))
        {
            sum = (sum*10 + p[i++]-'0')%d;
        }
        printf("Case %d: ",++t2);
        if(!sum)
            printf("divisible
");
        else
            printf("not divisible
");
    }
    return 0;
}