RT

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+1e4,M=1e3+1e2;
const ll Maxn=0x3ffffff,Minm=-0x3ffffff;
ll n;
ll a[N];
signed main()
{
    cin>>n;
    for(ll i=1;i<=n;i++)cin>>a[i];
    sort(a+1,a+n+1);
    for(ll i=1;i<=n;i++)
        if(a[i]!=a[i-1])cout<<a[i]<<" ";
}

退役后切的第一道水题…

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+1e4,M=1e3+1e2;
const ll Maxn=0x3ffffff,Minm=-0x3ffffff;
ll n;
ll a[N];
signed main()
{
    cin>>n;
    for(ll i=1;i<=n;i++)cin>>a[i];
    sort(a+1,a+n+1);
    for(ll i=1;i<=n;i++)
        if(a[i]!=a[i-1])//去重
            cout<<a[i]<<" ";
}

当我看题时，看到了至少，我以为这题很难的样子
可是我发现，那个无聊的人会把甘草拿过去然后再拿回来呀！！！！
直接算比平均数多的干草堆一共多了多少（（多的全拿走

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+1e4,M=1e3+1e2;
const ll Maxn=0x3ffffff,Minm=-0x3ffffff;
ll n;
ll a[N];
ll mid;
signed main()
{
    cin>>n;
    for(ll i=1;i<=n;i++)
    {
        cin>>a[i];
        mid+=a[i];
    }
    mid/=n;//均数
    ll ans=0;
    for(ll i=1;i<=n;i++)if(a[i]>mid)ans+=a[i]-mid;
    cout<<ans;
}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+1e4,M=1e3+1e2;
const ll Maxn=0x3ffffff,Minm=-0x3ffffff;
ll n;
ll a[N];
ll mid;
signed main()
{
    cin>>n;
    for(ll i=1;i<=n;i++)
    {
        cin>>a[i];
        mid+=a[i];
    }
    mid/=n;
    ll ans=0;
    for(ll i=1;i<=n;i++)if(a[i]>mid)ans+=a[i]-mid;
    cout<<ans;
}

这题不需要脑子吧

可是这道题让我想起了恶心的GDOI 2022 DAY2 T3 有兴趣的了解一下

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+1e4,M=1e3+1e2;
const ll Maxn=0x3fffffff,Minm=-0x3ffffff;
signed main()
{
    ll a,b,c;
    cin>>a>>b>>c;
    if((a<=11&&b<11)||(a<=11&&b<=11&&c<11))cout<<"-1";
    else cout<<(a-11)*1440+(b-11)*60+(c-11);
}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+1e4,M=1e3+1e2;
const ll Maxn=0x3fffffff,Minm=-0x3ffffff;
signed main()
{
    ll a,b,c;
    cin>>a>>b>>c;
    if((a<=11&&b<11)||(a<=11&&b<=11&&c<11))cout<<"-1";
    else cout<<(a-11)*1440+(b-11)*60+(c-11);
}