Large Division(大数取余模板)
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
模板题,记录一下
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 | #include<bits/stdc++.h> using namespace std; typedef long long ll; char a[1000]; int main() { int t,n,i; cin>>t; int h = 0; while (t--){ ll b,ans = 0; cin>>a>>b; if (b < 0){ b = -b; } for (i = 0; i < strlen(a); i++){ if (a[0] == '-' ){ a[i] = '0' ; } ans = (ans*10+(a[i]- '0' ))%b; } if (ans == 0){ printf( "Case %d: divisible\n" ,++h); } else { printf( "Case %d: not divisible\n" ,++h); } } return 0; } |
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