The 2025 ICPC Asia East Continent Online Contest (II)——C. Jiaxun! · znzryb

## 思路讲解

不难发现，可以用二分答案做。

那么思路就是发现ab，ac都可以看为是A的，但是因为ab，ac的不对称性，所以都要检验一遍。

具体check函数见下面。

```cpp
ll a,b,c,ab,ac,bc,abc;
bool check(ll mid){
    if(a+ac+ab+abc<mid) return false;
    if(b+bc+ab+abc<mid) return false;
    if(c+bc+ac+abc<mid) return false;
    if(a+c+ab+ac+bc+abc<2*mid) return false;
    if(a+b+ab+ac+bc+abc<2*mid) return false;
    if(b+c+ab+ac+bc+abc<2*mid) return false;
    if(N/3<mid) return false;
    return true;
}
```

## AC代码

[https://vjudge.net/solution/63690535](https://vjudge.net/solution/63690535)



::: 源代码

```cpp
// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define PF push_front
#define EB emplace_back
#define EF emplace_front
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define lson(o) (o<<1)
#define rson(o) (o<<1|1)
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())

using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    while(lT--)
        __();
    return 0;
}
/*

*/
ll a,b,c,ab,ac,bc,abc;
bool check(ll mid){
    if(a+ac+ab+abc<mid) return false;
    if(b+bc+ab+abc<mid) return false;
    if(c+bc+ac+abc<mid) return false;
    if(a+c+ab+ac+bc+abc<2*mid) return false;
    if(a+b+ab+ac+bc+abc<2*mid) return false;
    if(b+c+ab+ac+bc+abc<2*mid) return false;
    if(N/3<mid) return false;
    return true;
}
inline void __(){
    cin>>N;
    cin>>a>>b>>ab>>c>>ac>>bc>>abc;
    ll l=0,r=N;
    while (l<r) {
        ll mid=l+r+1>>1;
        if(check(mid)){
            l=mid;
        }else{
            r=mid-1;
        }
    }
    cout<<l<<"\n";
}
/*
AC
https://vjudge.net/solution/63690535
*/
```


:::


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