Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2269 Accepted Submission(s): 838
Problem Description
There is a number N.You should output "YES" if N is a multiple of 2, 3 or 5,otherwise output "NO".
Input
There are multiple test cases, no more than 1000 cases.
For each case,the line contains a integer N.(0<N<1030)
For each case,the line contains a integer N.(0<N<1030)
Output
For each test case,output the answer in a line.
Sample Input
2
3
5
7
Sample Output
YES
YES
YES
NO
Source
判断n是不是2,3,5的倍数
判断2,5,只要判断末尾是不是0,或者2,5的倍数,,所以只要对2,5求余,,二判断是不是3的倍数,只要所有的数加起来对3求余就好
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 #include <string> 5 #include <string.h> 6 #include <cstring> 7 #include <stdio.h> 8 #define lowbit(x) x&(-x) 9 10 using namespace std; 11 const int maxn=1e5+10; 12 typedef long long ll; 13 14 int main(){ 15 ios::sync_with_stdio(0); 16 string s; 17 while(cin>>s){ 18 int len=s.length(); 19 int sum=0; 20 for(int i=0;i<s.length();i++){ 21 sum+=s[i]-'0'; 22 } 23 if(sum%3==0||((s[len-1]-'0')%2==0)||((s[len-1]-'0')%5==0)){ 24 cout<<"YES"<<endl; 25 } 26 else { 27 cout<<"NO"<<endl; 28 } 29 } 30 return 0; 31 }