#include <iostream>
using namespace std;
int main()
{cout<<"51"<<endl;return 0;
}

假设黄水晶的个数是$x$,那蓝水晶个数肯定是$x-1$,红水晶是$x-2$,总的水晶数是$3x-3$。总的水晶数不能超过埚的容量所以$a_i>=3x-3$枚举一下即可时间复杂度$O(n{a}_i)$

void solve()
{cin>>n;for(int i=1;i<=n;i++)cin>>a[i];int res=0;for(int i=1;i<=n;i++){for(int j=a[i];j>=0;j--){if(3*j-3<=a[i]&&j-2>=0){res+=j-2;break;}}}cout<<res<<endl;
}

贪心，只要正数就用+号不够就用-号，负数用-号不够用+号即可

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+4;
typedef long long ll;
pair<int,int> p[N];
bool cmp(pair<int,int> x,pair<int,int> y)
{return x.first>y.first;
}
int main()
{int n,a,b;cin>>n>>a>>b;for (int i=1;i<=n;++i){int x;cin>>x;p[i].first=abs(x);p[i].second=x;}ll sum=p[1].second;sort(p+2,p+1+n,cmp);for (int i=2;i<=n;++i){if (p[i].second<=0){if (b>=1) sum+=(ll)p[i].first,b--;else sum+=(ll)p[i].second,a--;} else{if (a>=1) sum+=(ll)p[i].first,a--;else sum-=(ll)p[i].first,b--;}}cout<<sum;return 0;
}

暴力枚举每一个点是否有v然后取$max$即可

int m,n;
char g[N][N];
int cnt;void calc(int x,int y)
{for(int i=0;i<=max(n,m);i++){int tx1=x-i,ty1=y-i;int tx2=x-i,ty2=y+i;if(tx1<=0||ty1<=0||ty1>m||tx1>n||tx2<=0||ty2<=0||ty2>m||tx2>n)return;if(tx1==x&&ty1==y)continue;if(g[tx1][ty1]=='.'||g[tx2][ty2]=='.')return;cnt++;   }
}void solve()
{cin>>n>>m;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)cin>>g[i][j];int res=0;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(g[i][j]=='*'){cnt=0;calc(i,j);res=max(res,cnt);}}cout<<res<<endl;
}

贪心，看数据范围都是正数，那肯定要取$ma{x}_{i=1}^n{a}_i$ 然后就遍历一下找出最大的即可

void solve()
{cin>>n;for(int i=1;i<=n;i++)cin>>a[i];sort(a+1,a+1+n);reverse(a+1,a+1+n);int res=a[1]+a[2];string t1=to_string(a[1]);string t2=to_string(a[2]);reverse(t1.begin(),t1.end());reverse(t2.begin(),t2.end());int v=0;for(int i=0;i<min(t1.size(),t2.size());i++){if(t1[i]==t2[i])v+=t1[i]-'0';}res-=v;int s=a[1];for(int i=2;i<=n;i++){string t3=to_string(a[i]);reverse(t3.begin(),t3.end());int tt=0;for(int i=0;i<min(t1.size(),t3.size());i++){if(t1[i]==t3[i])tt+=t3[i]-'0';}res=max(res,s+a[i]-tt);}cout<<res<<endl;
}