52 条题解
-
-1
hetao36625371 LV 9 @ 2024-2-22 17:31:01
兄弟们，这道题简直太太太简单了，话不多说上代码，绝对AC，没AC我吃屎！！！``` #include <iostream> using namespace std; int main() { int n; cin>>n; if(n%20) { cout<<"2"<<endl; } if(n%30) { cout<<"3"<<endl; } if(n%50) { cout<<"5"<<endl; } if(n%70) { cout<<"7"<<endl; } return 0; }
-
hetao11417753
@ 2024-2-23 9:50:43
你这样我们怎么复制？
-
-
-1
hetao216752
LV 7
@ 2024-1-1 13:12:45
#include<bits/stdc++.h>//AC using namespace std; int main() { int n; cin>>n; if(n%2==0) { cout<<2<<"\n"; } if(n%3==0) { cout<<3<<"\n"; } if(n%5==0) { cout<<5<<"\n"; } if(n%7==0) { cout<<7<<"\n"; } return 0; } -
-1
hetao3937345
LV 10
@ 2023-11-12 14:58:50
这道题非常简单，用四个分支if语句即可完成，请看题解：
#include <iostream> using namespace std; int main() { int n; cin >> n; if (n % 2 == 0) { cout << 2 << endl; } if (n % 3 == 0) { cout << 3 << endl; } if (n % 5 == 0) { cout << 5 << endl; } if (n % 7 == 0) { cout << 7 << endl; } return 0; }（已AC，请放心食用( •̀ ω •́ )✧~） 最后，请大家~~点个赞再走呗~~，谢谢！
-
-1
hetao15155991 LV 8 @ 2023-10-28 21:31:34
没难度呀
-
hetao28428387 @ 2024-2-2 14:54:51
没难度你来发呀
#include<iostream> using namespace std; int main() { int a; cin>>a; if (a%2==0) { cout<<2<<endl; } for (int i=3;i<=7;i+=2) { if (a%i==0) { cout<<i<<endl; } } } -
再打一遍
#include<iostream> using namespace std; int main() { int a; cin>>a; if (a%2==0) { cout<<2<<endl; } for (int i=3;i<=7;i+=2) { if (a%i==0) { cout<<i<<endl; } } return 0; }
-
-
-1
hetao11069165 LV 6 @ 2023-10-4 13:10:06
较为简单(或是灰常简单)的if分支问题呦!挨个除遍就行啦!上代码: #include <iostream> using namespace std; int main() { int n; cin >> n; if (n % 2 == 0) { cout << "2" << endl; } if (n % 3 == 0) { cout << "3" << endl; } if (n % 5 == 0)**** { cout << "5" << endl; } if (n % 7 == 0) { cout << "7" << endl; } return 0; } 点赞再抱走哈!AC在N秒(N <= 30)后与你会面(除非网络不给力……)
-
-1
hetao1371772 LV 7 @ 2023-10-4 12:10:49
#include <iostream> using namespace std; int main() { int n; cin>>n; if (n%2==0)//判断是否能被2整除 { cout<<"2"<<endl;//输出2 } if (n%3==0)//判断是否能被3整除 { cout<<"3"<<endl;//输出3 } if (n%5==0)//判断是否能被5整除 { cout<<"5"<<endl;//输出5 } if (n%7==0)//判断是否能被7整除 { cout<<"7";//输出7 } return 0; } -
-1
hetao10449306 LV 6 @ 2023-10-2 21:33:16
#include<bits/stdtr1c++.h> using namespace std; int main() { int n; cin>>n; if(n%2==0){ cout<<2<<endl; } if(n%3==0){ cout<<3<<endl; } if(n%5==0){ cout<<5<<endl; } if(n%7==0){ cout<<7<<endl; } return 0; } -
-1
#include<bits/stdtr1c++.h> using namespace std; int main() { int n; cin>>n; if(n%2==0){ cout<<2<<endl; } if(n%3==0){ cout<<3<<endl; } if(n%5==0){ cout<<5<<endl; } if(n%7==0){ cout<<7<<endl; } return 0; } -
-1
#include<bits/stdtr1c++.h> using namespace std; int main() { int n; cin>>n; if(n%2==0){ cout<<2<<endl; } if(n%3==0){ cout<<3<<endl; } if(n%5==0){ cout<<5<<endl; } if(n%7==0){ cout<<7<<endl; } return 0; } -
-1
hetao11666439
LV 10
@ 2023-9-19 21:35:32
#include<bits/stdc++.h> using namespace std; int main() { int c; cin >> c; if(c % 2 == 0) { cout << "2" <<endl; } if(c % 3 == 0) { cout << "3"<<endl; } if(c % 5 == 0) { cout << "5"<<endl; } if(c % 7 == 0) { cout << "7"<<endl; } return 0; }//已AC -
-1
hetao8304638 LV 8 @ 2023-8-23 9:36:03
#include <iostream> using namespace std; int main() { int n; cin>>n; if (n%2==0)//判断是否能被2整除 { cout<<"2"<<endl;//输出2 } if (n%3==0)//判断是否能被3整除 { cout<<"3"<<endl;//输出3 } if (n%5==0)//判断是否能被5整除 { cout<<"5"<<endl;//输出5 } if (n%7==0)//判断是否能被7整除 { cout<<"7";//输出7 } return 0; }编程不易😕 ，赞在哪里👀️ ？
wsy1998
LV 10
MOD
@ 2022-2-26 20:20:25
