hdu 5585 Numbers
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
读入%s字符串
2和5看大数的最后一位,3就把所有的位加起来看看是不是3的倍数
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 #include <stack> 15 using namespace std; 16 #define PI acos(-1.0) 17 #define max(a,b) (a) > (b) ? (a) : (b) 18 #define min(a,b) (a) < (b) ? (a) : (b) 19 #define ll long long 20 #define eps 1e-10 21 #define MOD 1000000007 22 #define N 1000000 23 #define inf 1e12 24 char s[36]; 25 int main() 26 { 27 28 29 while(scanf("%s",s)!=EOF){ 30 int len=strlen(s); 31 //printf("len=%d\n",len); 32 int sum=0; 33 for(int i=0;i<len;i++){ 34 sum=sum+s[i]-'0'; 35 } 36 //printf("sum=%d\n",sum); 37 if(sum%3==0){ 38 printf("YES\n"); 39 continue; 40 } 41 int ans=s[len-1]-'0'; 42 //printf("ans=%d\n",ans); 43 if((ans==0) || (ans==2) || (ans==4) || (ans==6) || (ans==8) || (ans==5)){ 44 printf("YES\n"); 45 continue; 46 } 47 printf("NO\n"); 48 } 49 50 51 return 0; 52 }