思路:一个大数对一个数取余,可以把大数看成各位数的权值与个位数乘积的和。比如1234=( (1*10 + 2 )* 10 +3 )* 10+4,对这个数进行取余运算就是上面基本加和乘的应用。
代码实现:
int len = a.length();
int ans=0;
for( int i=0; i<len; i++){
ans = ( ans * 10 + a[i] - '0' ) mod b;
}
举个LightOj的例子,代码实现如下:
#include<bits/stdc++.h>
using namespace std;
int main(){
char s[100050];
long long b,res=1;
int t;
cin>>t;
while(t--){
long long ans=0;
cin>>s>>b;
int l=strlen(s);
for(int i=0;i<l;i++){
if(s[i]=='-')
continue;
ans=(ans*10+s[i]-'0')%b;
}
//printf("Case %d: ",res++);
cout<<"Case "<<res++<<": ";
if(ans)
cout<<"not divisible"<<endl;
else
cout<<"divisible"<<endl;
}
return 0;
}
ps:我想带你一起去看海!
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