| 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;
}