E.cpp

#include <bits/stdc++.h>

#define fi first
#define se second
#define db double
#define U unsigned
#define P std::pair<int,int>
#define LL long long
#define pb push_back
#define MP std::make_pair
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#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(x) std::cerr << #x << '=' << x << std::endl
#define int LL
LL a[4],b[4],n;
const int MAXN = 2000+5;

namespace Flow{
    struct Edge{
        int to,w,nxt;
    }e[MAXN<<1];
    int head[MAXN],cur[MAXN],cnt=1;
    int dep[MAXN];
    int S,T,N;

    inline void add(int u,int v,int w){
        e[++cnt] = (Edge){v,w,head[u]};head[u] = cnt;
        e[++cnt] = (Edge){u,0,head[v]};head[v] = cnt;
    }

    inline bool bfs(){
        FOR(i,0,N) cur[i] = head[i],dep[i] = 0;
        std::queue<int> q;q.push(S);dep[S] = 1;
        while(!q.empty()){
            int v = q.front();q.pop();
            for(int i = head[v];i;i = e[i].nxt){
                if(e[i].w > 0 && !dep[e[i].to]){
                    dep[e[i].to] = dep[v] + 1;
                    if(e[i].to == T) return true;
                    q.push(e[i].to);
                }
            }
        }
        return dep[T];
    }

    inline int dfs(int v,int flow=1e9){
        if(v == T) return flow;
        if(!flow) return 0;
        int ans = 0;
        for(int &i = cur[v];i;i = e[i].nxt){
            if(e[i].w > 0 && dep[e[i].to] == dep[v] + 1){
                int t = dfs(e[i].to,std::min(flow,e[i].w));
                if(t>0){
                    ans += t;flow -= t;
                    e[i].w -= t;e[i^1].w += t;
                    if(!flow) break;
                }
            }
        }
        return ans;
    }

    inline int Dinic(){
        int ans = 0,t = 0;
        while(bfs()) while((t=dfs(S))) ans += t;
        return ans;
    }
};
using namespace Flow;

signed main(){
	scanf("%lld",&n);
	FOR(i,1,3) scanf("%lld",a+i);
	FOR(i,1,3) scanf("%lld",b+i);
	LL a1 = 0,a2 = 0;
	FOR(i,1,3) a2 += std::min(a[i],b[i%3+1]);
	N = 8;S = 7;T = 8;
	FOR(i,1,3) add(S,i,a[i]),add(i+3,T,b[i]),add(i,i+3,3e9);
	add(1,3+3,3e9);add(2,1+3,3e9);add(3,2+3,3e9);
	a1 = Dinic();
	printf("%lld %lld\n",n-a1,a2);
	return 0;
}