1 //检验一个大数是否能3整除 2 //A number is divisible by 3 if sum of its digits is divisible by 3. 3 //we cannot use n % 3 to check if a number is divisible by 3 or not. 4 //Remainder of 10i divided by 3 is 1 So powers of 10 only result in value 1. 5 #include<bits/stdc++.h> 6 using namespace std; 7 8 int main() 9 { 10 string s; 11 cin>>s; 12 int len=s.length(); 13 int sum=0; 14 for(int i=0;i<len;i++) 15 { 16 sum+=s[i]-‘0‘; 17 } 18 if(sum%3==0) 19 cout<<"yes"<<endl; 20 else 21 cout<<"no"<<endl; 22 return 0; 23 }
1 # “脱数“代码 2 def check(num): 3 while num>0: 4 rem=num%10 5 sum+=rem 6 num/=10
原文地址:https://www.cnblogs.com/chuanwen-tech/p/11296178.html
时间: 2024-10-29 19:52:00