E. Rock, Paper, Scissors

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Alice and Bob have decided to play the game "Rock, Paper, Scissors".

The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:

  • if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
  • if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
  • if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.

Alice and Bob decided to play exactly nn rounds of the game described above. Alice decided to show rock a1a1 times, show scissors a2a2 times and show paper a3a3 times. Bob decided to show rock b1b1 times, show scissors b2b2 times and show paper b3b3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a1+a2+a3=na1+a2+a3=n and b1+b2+b3=nb1+b2+b3=n.

Your task is to find two numbers:

  1. the minimum number of round Alice can win;
  2. the maximum number of rounds Alice can win.

Input

The first line of the input contains one integer nn (1≤n≤1091≤n≤109) — the number of rounds.

The second line of the input contains three integers a1,a2,a3a1,a2,a3 (0≤ai≤n0≤ai≤n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a1+a2+a3=na1+a2+a3=n.

The third line of the input contains three integers b1,b2,b3b1,b2,b3 (0≤bj≤n0≤bj≤n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b1+b2+b3=nb1+b2+b3=n.

Output

Print two integers: the minimum and the maximum number of rounds Alice can win.

Examples

input

2
0 1 1
1 1 0

output

0 1

input

15
5 5 5
5 5 5

output

0 15

input

3
0 0 3
3 0 0

output

3 3

input

686
479 178 29
11 145 530

output

22 334

input

319
10 53 256
182 103 34

output

119 226

Note

In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.

In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.

In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.

题意:

alice和bob玩剪刀石头布,已知alice出了a1个石头、a2个剪刀、a3个步,bob出了b1个石头、b2个剪刀、b3个布,问alice最少赢了多少次,最多赢了多少次。

思路:

1、最多赢多少次

让alice尽量赢就可以了 maxx = min(a1, b2) + min(a2, b3) + min(a3, b1);

2、最少赢多少次

考虑让alice尽量输或平,也就是让a1和b1、b3更多地抵消掉,a2和b2、b1更多地抵消掉,a3和b3、b2更多地抵消掉。抵消后,最优的最终状态肯定是(全为0)或者(alice剩下一种且bob剩下一种),全为0就不用说了,剩下的情况肯定都是alice赢(因为抵消的都是对赢没有贡献的,无法抵消肯定是有贡献,也就是alice赢)。

还有一个小问题,alice如果有剩余一定是一种吗,答案是肯定的,因为如果alice剩下了两种:a2个剪刀和a3个布,bob剩下了一种:b1个石头,(a2 + a3 == b1就不用解释了8)此时还可以让alice的剪刀和石头相抵消,也就是某人剩下两种时根本不是最终状态,还可以继续抵消。因此有剩余情况下的最终状态肯定是每人一种。

下面来讨论一下抵消后alice有剩余的情况:

(1)如果剩下a1,那对应剩下的就是b2,a1 == b2。

整个过程让b2尽量消失,所以a1、b2的剩余量均为 b2 - a2 - a3。(这里也可以考虑成让a1尽量地消失,a1、b2的剩余量为a1 - b1 - b3)

(2)如果剩下a2,那对应剩下的就是b3,a2 == b3。

整个过程让b3尽量消失,所以a2、b3的剩余量均为 b3 - a1 - a3。(这里也可以考虑成让a2尽量地消失,a2、b3的剩余量为a2 - b1 - b2)

(3)如果剩下a3,那对应剩下的就是b1,a3 == b1。

整个过程让b1尽量消失,所以a3、b1的剩余量均为 b1 - a1 - a2。(这里也可以考虑成让a3尽量地消失,a3、b1的剩余量为a3 - b2 - b3)

现在来看一下这三个数(b2 - a2 - a3)、(b3 - a1 - a3)、(b1 - a1 - a2)这三个数中最多只有一个正数,证明如下:

假设 b2 - a2 - a3 > 0,即 b2 > a2 + a3

因为 b1 + b2 + b3 == a1 + a2 + a3 == n

所以 b1 + b3 < a1

所以 b1 < a1 + a2 且 b3 < a1 + a3

所以上面三个数中最多只有一个正数。

所以最终的答案就是从(b2 - a2 - a3)、(b3 - a1 - a3)、(b1 - a1 - a2)中选一个最大值就可以了。

如果上面考虑让a1、a2、a3尽量地消失就是从(a1 - b1 - b3)、(a2 - b1 - b2)、(a3 - b2 - b3)中选一个最大值,答案是一样的,因为alice和bob剩下的数肯定相同。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 1e4 + 10;
const int M = 1e5 + 10;
 
int main() {
    int n, a[4], b[4];
    scanf("%d", &n);
    for(int i = 1; i <= 3; ++i) scanf("%d", &a[i]);
    for(int i = 1; i <= 3; ++i) scanf("%d", &b[i]);
    ///surprise  max居然可以加大括号
    int minn = max({0, a[1] - b[1] - b[3], a[2] - b[2] - b[1], a[3] - b[3] - b[2]});
    int maxx = min(a[1], b[2]) + min(a[2], b[3]) + min(a[3], b[1]);
    printf("%d %d\n", minn, maxx);
	return 0;
}

最小费用最大流:

超级源点向alice的石头、剪刀、布连边,流量分别为a1、a2、a3,费用都为0

bob的石头、剪刀、布连边,流量分别为b1、b2、b3,费用都是0

alice的三种决策向bob的三种决策连边,流量都为inf,只有alice赢时费用为1(贡献一个赢)

套板子就完了
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define P pair<int,int>
const int inf = 0x3f3f3f3f;
const int N = 1e4 + 10;
const int M = 1e5 + 10;
 
struct node {
    int x, y;
};
vector<node>peo, house;
 
struct Edge {
	int to, cap, cost, rev;
};
 
int n, m, s, t;
int u, v, c, w;
int maxFlow, minCost;
 
vector<Edge> G[N];
int h[N];   ///顶点的势
int dist[N], prevv[N], preve[N]; ///dist最短距离 prevv最短路中的父节点 preve最短路中的父边
 
void init() {
    maxFlow = 0;
    minCost = 0;
    for(int i = 0; i < N; ++i)
        G[i].clear();
}
 
void addedge(int from, int to, int cap, int cost) {
	Edge temp1 = { to, cap, cost, (int)G[to].size() };
	Edge temp2 = { from, 0, -cost, (int)G[from].size() - 1 };
	G[from].push_back(temp1);
	G[to].push_back(temp2);
}
 
void MCMF(int s, int t, int f, int n) {    //n是总结点数
	fill(h + 1, h + 1 + n, 0);
	while (f > 0) {
		priority_queue<P, vector<P>, greater<P> > D;
		memset(dist, inf, sizeof(dist));
		dist[s] = 0; D.push(P(0, s));
		while (!D.empty()) {
			P now = D.top(); D.pop();
			if (dist[now.second] < now.first) continue;
			int v = now.second;
			for (int i = 0; i<(int)G[v].size(); ++i) {
				Edge &e = G[v][i];
				if (e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
					dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
					prevv[e.to] = v;
					preve[e.to] = i;
					D.push(P(dist[e.to], e.to));
				}
			}
		}
		if (dist[t] == inf) break;
		for (int i = 1; i <= n; ++i) h[i] += dist[i];
		int d = f;
		for (int v = t; v != s; v = prevv[v])
			d = min(d, G[prevv[v]][preve[v]].cap);
		f -= d; maxFlow += d;
		minCost += d * h[t];
		for (int v = t; v != s; v = prevv[v]) {
			Edge &e = G[prevv[v]][preve[v]];
			e.cap -= d;
			G[v][e.rev].cap += d;
		}
	}
}
 
int main() {
    int cnt, a[4], b[4];
    while(~scanf("%d", &cnt)) {
        init();
        n = 8, s = 0, t = 7;
        for(int i = 1; i <= 3; ++i) {
            scanf("%d", &a[i]);
            addedge(s, i, a[i], 0);
        }
        for(int i = 1; i <= 3; ++i) {
            scanf("%d", &b[i]);
            addedge(i + 3, t, b[i], 0);
        }
        int maxx = min(a[1], b[2]) + min(a[2], b[3]) + min(a[3], b[1]);
        for(int i = 1; i <= 3; ++i) {
            for(int j = 1; j <= 3; ++j) {
                int cst = 0;
                if(i == 1 && j == 2) cst = 1;
                if(i == 2 && j == 3) cst = 1;
                if(i == 3 && j == 1) cst = 1;
                addedge(i, j + 3, inf, cst);
            }
        }
        MCMF(s, t, inf, n + m + 2);
        printf("%d %d\n", minCost, maxx);
    }
	return 0;
}
/*
2
0 1 1
1 1 0
15
5 5 5
5 5 5
3
0 0 3
3 0 0
686
479 178 29
11 145 530
319
10 53 256
182 103 34
94317
66277 24448 3592
3499 24653 66165
*/