题目:LightOJ :: Problem 1214 - Large Division
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:
给一个大数a判断a是否整除b(注意a的正负)
数论原理:
(a + b) % p = (a % p + b % p) % p
同余模定理:https://baike.baidu.com/item/%E5%90%8C%E4%BD%99%E5%AE%9A%E7%90%86
同余模定理应用:https://baike.baidu.com/item/%E5%8F%96%E6%A8%A1%E8%BF%90%E7%AE%97/10739384
代码:
#include<bits/stdc++.h> using namespace std; int main() { char a[205]; int n; int b; cin>>n; for(int i=1;i<=n;i++) { cin>>a>>b; int len=strlen(a); int flag=0,ans=0; if(a[0]=='-') flag=1; for(int j=flag;j<len;j++) { ans=(int)(((long long)ans*10+(a[j]-'0'))%b); } printf("Case %d: ",i); if(ans==0) printf("divisible\n"); else printf("not divisible\n"); } return 0; }