Light OJ 1214 - Large Division 【同余定理】

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

#include<cstring>
#include<cstdio>
char a[2200000];
int b;
int main(){
int T,t;
scanf("%d", &T);
for(t = 1; t <= T;t++)
{
scanf("%s", a);
scanf("%d", &b);
int len = strlen(a);
long long sum = 0;//int  就wa了.....
for(int i = 0; i < len; i++)
{
if(a[i]=='-')	continue;
sum = (sum*10+a[i]-'0')%b;
}
if(sum==0)
printf("Case %d: divisible\n", t);
else
printf("Case %d: not divisible\n", t);
}
return 0;
}