[C++知识库]纸牌游戏（贪心）

// Problem: 纸牌游戏
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1039
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 2022-03-04 20:29:56
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
const int N= 2e5 + 10;

ll a[N], b[N];
int n;
void solve() {
	cin >> n;
    for (int i = 1; i <= n; i++)
            cin >> a[i];
    for (int i = 1; i <= n ;i ++)
            cin >> b[i];
    sort(b + 1,b + 1 + n, greater<ll>());
    sort(a + 1, a + 1 +n);
    ll ans = 0;
    for (int i = 1; i <= n; i ++)
            ans += max ((ll)0, b[i] - a[i]);
   cout << ans << endl;
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

我认为a数组是不能变的实际上我们可以让a与任何一个b中的数配对，说明a也是任意排的。所以我们可以让 最大和最小配对。这样的贡献最大。从贡献的角度思考