```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int a,b,c,s1=10000,s2=1000;//刚开始s1 s2要尽量大
cin>>a>>b>>c;
if(abs(a-b-c)<s1-s2){//接下来哪个差距比较小就选哪个
s1=max(a,b+c);
s2=min(a,b+c);
}
if(abs(b-a-c)<s1-s2){
s1=max(b,a+c);
s2=min(b,a+c);
}
if(abs(c-b-a)<s1-s2){
s1=max(c,b+a);
s2=min(c,b+a);
}
cout<<s1<<' '<<s2;
return 0;
}
```
by LittleL0124 @ 2024-04-27 16:18:36
@[hc_20120919](/user/1235710) 不太能看懂你的做法,给你说个思路:设a,b,c中最大的为n,和为m,那么z拿n,m-n中大的,yz拿小的
by lczcy1 @ 2024-04-27 16:19:07
不懂你的做法
by shimucheng @ 2024-04-27 16:23:47
按照你的做法,min和max是c++的函数,最好别用
by shimucheng @ 2024-04-27 16:41:51
思路:先把三盒加起来(ans),如是偶数,cout<<ans/2+1<<" "<<ans/2-1,如是奇数,cout<<ans/2+1<<" "<<ans/2
by shimucheng @ 2024-04-27 16:46:24