36分,求助。

B2043 判断能否被 3,5,7 整除

@[liuliuzi](/user/1430545) 布什戈门你贴一下代码阿
by T_TwY @ 2024-08-26 18:41:08


```cpp #include<bits/stdc++.h> using namespace std; int main(){ int a; cin>>a; if (a%105==0){ cout<<"3 5 7"; }else if (a%35==0){ cout<<"5 7"; }else if (a%21==0){ cout<<"3 7"; }else if (a%15==0){ cout<<"3 5"; }else if (a%3==0){ cout<<"3"; }else if (a%5==0){ cout<<"5"; }else if (a%7==0){ cout<<"7"; }else{ cout<<"n"; } return 0; } ``` @[liuliuzi](/user/1430545) 你原来的码风有点不好,我改了一下 另外下次记得把代码放到帖子里
by Qiu101029 @ 2024-08-26 18:43:19


@[liuliuzi](https://www.luogu.com.cn/user/1430545) 哥们,贴下代码呗~~~
by zhanghengrui0502 @ 2024-08-26 18:44:25


~~尽管改完之后还是不太好~~
by Qiu101029 @ 2024-08-26 18:44:35


@[Qiu101029](/user/1200904) 其实可以把if并列的,会更简洁,最后加一个bool或者int来判断是否为n就行
by jiangsm @ 2024-08-26 18:45:14


@[jiangsm](/user/1386287) 我的代码如下 ```cpp #include <bits/stdc++.h> using namespace std; int main() { long long a,t=0; cin>>a; if(a%3==0) { cout<<3<<" "; t++; } if(a%5==0) { cout<<5<<" "; t++; } if(a%7==0) { cout<<7; t++; } if(t==0) { cout<<"n"; } return 0; } ``` ~~不知道为什么没有人发这种类型的题解~~
by jiangsm @ 2024-08-26 18:46:31


@[jiangsm](/user/1386287) 我知道,但他这么写的
by Qiu101029 @ 2024-08-26 18:46:34


@[Qiu101029](/user/1200904) 行
by jiangsm @ 2024-08-26 18:47:05


不是哥们你代码呢???
by zhangyiquan @ 2024-08-28 10:21:36


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