题目:
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
6
101 101
0 67
-101 101
7678123668327637674887634 101
11010000000000000000 256
-202202202202000202202202 -101
Sample Output
Case 1: divisible
Case 2: divisible
Case 3: divisible
Case 4: not divisible
Case 5: divisible
Case 6: divisible
题意:
给你两个数a,b,让你求出来a是否能够被b整除。
思路:
需要注意的是数字a太大了,所以要用数组来存储,同时还要注意数字b可能超出了int范围,要用long long int,同时也要注意一些细这道题不难;
代码如下:
#include<stdio.h>
#include<string.h>
#define N 100010
int t;
char a[N];
long long int b,s;//注意范围;
int main()
{
int kk=1;
scanf("%d",&t);
while(t--)
{
memset(a,0,sizeof 0);
scanf("%s%lld",a,&b);//注意数字b的范围;
int l=strlen(a);
s=0;
if(b<0)//注意正负数;
b=-b;
for(int i=0; i<l; i++)
{
if(a[i]=='-')//注意正负数;
continue;
s=(s*10+a[i]-'0')%b;
}
if(s==0)
printf("Case %d: divisible\n",kk++);
else
printf("Case %d: not divisible\n",kk++);
}
return 0;
}