1 solutions
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0
C :
#include<stdio.h> void main(){ int i,n; scanf("%d",&n); i=1; while(i<=n){ if(i%2==0&&i%3!=0){ printf("%d\n",i); } i++; } }
C++ :
#include <iostream> using namespace std; int main(){ /* 第一步:输出1~n的每个数 第二步:判断该数是否满足条件(是2的倍数但非3的倍数) */ int i,n; cin>>n; i = 1; while(i <= n){ if(i % 2 == 0 && i % 3 != 0){ cout<<i<<endl; } i++; } }
Java :
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); for(int i=1;i<=n;i++) { if(i%2==0 && i%3!=0) { System.out.println(i); } } } }
Python :
n = int(input()) for i in range(1, n + 1): if i % 2 == 0 and i % 3 != 0: print(i)
- 1
Information
- ID
- 10495
- Time
- 1000ms
- Memory
- 64MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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