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  • hdu 5585

    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)
     
    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 }
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  • 原文地址:https://www.cnblogs.com/z-712/p/8733903.html
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