The 19th Heilongjiang Provincial Collegiate Programming Contest

1|0B. String

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

using vi = vector<int>;


i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	string s;
	cin >> s;
	vector<char> t;
	for(auto c : s){
		t.push_back(c);
		if(t.size() >= 3){
			if(t[t.size() - 1] == t[t.size() - 2] and t[t.size() - 1] == t[t.size() - 3])
				t.pop_back(),t.pop_back(),t.pop_back();
		}
	}
	if(t.empty()) cout << "NAN\n";
	else {
		for(auto c : t)
			cout << c;
		cout << "\n";
	}
	return 0;
}

2|0D. Card Game

贪心题，尽可能的给对方造成高的伤害。

#include <bits/stdc++.h>

using namespace std;

#define ll long long

void solve(){
	int n,hpa,hpb;
	cin >> n >> hpa >> hpb;

	vector<int> a(n),b(n);
	for (int i = 0;i < n;i++){
		cin >> a[i];
	}
	for (int i = 0;i < n;i++){
		cin >> b[i];
	}

	sort(a.begin(),a.end(),greater<>());
	sort(b.begin(),b.end(),[](int x,int y){
		if (x != -1 && y != -1){
			return x < y;
		}else{
			return (x == -1) < (y == -1);
		}
	});

	for (int i = 0;i < n && hpa > 0 && hpb > 0;i++){
		if (a[i] == -1 || b[i] == -1) continue;
		hpa -= b[i];
		hpb -= a[i];
	}
	if(hpb <= 0 and hpa > 0) cout << "yes\n";
	else cout << "no\n";
	return;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int t;
	cin >> t;

	while (t--){
		solve();
	}

	return 0;
}

3|0F. Photography

首先在枚举中间的三个点，然后贪心的选择首尾两个点。

#include <bits/stdc++.h>

using namespace std;

#define ll long long

const int inf = 1e9;
const int N = 5e3 + 7;
int n,m,a[N],d[N][6],vis[N],ans = 0;
vector<int> e[N];

using pii = pair<int,int>;

void dfs(int fa,int u,int t,int res){
	if (t > 5) return;
 
	d[fa][t] = max(d[fa][t],res);
	ans = max(d[fa][t],ans);
	for (auto v : e[u]){
		if (vis[v]) continue;
		vis[v] = 1;
		dfs(fa,v,t + 1,res + a[v]);
		vis[v] = 0;
	}
}

int main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);

	cin >> n >> m;
	for (int i = 1;i <= n;i++){
		cin >> a[i];
	}
	for (int i = 1;i <= m;i++){
		int u,v;
		cin >> u >> v;
		e[u].push_back(v);
		e[v].push_back(u);
	}
	vector<array<int,3>> edge;

	for(int i = 1; i <= n; i ++){
		for(int x = 0; x < e[i].size(); x ++){
			for(int y = 0; y < x; y ++ ){
				edge.push_back({e[i][x], i , e[i][y]});
			}
		}
	}

	if(edge.empty()) {
		for (int i = 1;i <= n;i++){
			vis[i] = 1;
			dfs(i,i,1,a[i]);
			vis[i] = 0;
		}
		cout << ans << "\n";
		return 0;
	}

	vector<set<pii>> cnt(n + 1);

	for(int x = 1; x <= n; x ++){
		for(const auto y : e[x]){
			cnt[x].emplace(-a[y], y);
			if(cnt[x].size() > 5) 
				cnt[x].erase(prev(cnt[x].end()));
		}
	}
	for(const auto &[x,y,z] : edge){
		set<int> p;
		p.insert(x);
		p.insert(y);
		p.insert(z);
		for(auto [_ , i] : cnt[x]){
			if(p.count(i)) continue;
			p.insert(i);
			break;
		}
		for(auto [_ , i] : cnt[z]){
			if(p.count(i)) continue;
			p.insert(i);
			break;
		}
		int cur = 0;
		for(auto i : p)
			cur += a[i];
		ans = max(ans, cur);
	}
	cout << ans << "\n";
	return 0;
}

4|0G. Grey-like Code

打表找规律，发现答案是 $2^{2^{n-1} - n}$

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

using vi = vector<int>;

#define int i64

const int mod = 998244353;

int power(int x, int y , int p = mod) {
	int ans = 1;
	while(y) {
		if(y & 1) ans = ans * x % p;
		x = x * x % p , y /= 2;
	}
	return ans;
}

i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int n;
	cin >> n;
	int phi = mod - 1;
	int x = (power(2 , n - 1 , phi) - n + phi) % phi;
	cout << power(2 , x);
	return 0;
}

5|0I. This is an easy problem

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

using vi = vector<int>;


#define lowbit(x) (x & -x)

i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int x , res = 0;
	cin >> x;
	while(x)
		x -= lowbit(x), res ++;
	cout << res << "\n"; 
	return 0;
}

6|0J. Trade

动态规划，记 $f[i][j]$ 表示从 $(1,1)$ 到 $(i,j)$ 只经过合法路径所需要的最短距离。

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;

#define int i64

using vi = vector<int>;

const int inf = 1e18;

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector a(n + 1, vi(m + 1)), b(n + 1, vi(m + 1));
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> b[i][j];
    vector f(n + 1, vi(m + 1, inf));
    for (int i = 1, g; i <= n; i++)
        for (int j = 1; j <= m; j++) {
            if (i == 1 and j == 1) f[1][1] = b[1][1];
            else {
                g = min(f[i - 1][j], f[i][j - 1]) + b[i][j];
                if (g + a[1][1] <= a[i][j]) f[i][j] = g;
            }
        }

    for (int i = 1; i <= m; i++)
        if (f[n][i] != inf) {
            cout << "YES\n";
            return 0;
        }
    for (int i = 1; i <= n; i++)
        if (f[i][m] != inf) {
            cout << "YES\n";
            return 0;
        }
    cout << "NO\n";
    return 0;
}

7|0K. Puzzle

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

using vi = vector<int>;


#define lowbit(x) (x & -x)

i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	vi a(4);
	for(auto & i : a) cin >> i;
	sort(a.begin(), a.end());
	set<int> cnt;
	do{
		cnt.insert(a[0]+a[1]+a[2]+a[3]);
		cnt.insert(a[0]+a[1]+a[2]-a[3]);
		cnt.insert(a[0]+a[1]+a[2]*a[3]);
		cnt.insert(a[0]+a[1]-a[2]+a[3]);
		cnt.insert(a[0]+a[1]-a[2]-a[3]);
		cnt.insert(a[0]+a[1]-a[2]*a[3]);
		cnt.insert(a[0]+a[1]*a[2]+a[3]);
		cnt.insert(a[0]+a[1]*a[2]-a[3]);
		cnt.insert(a[0]+a[1]*a[2]*a[3]);

		cnt.insert(a[0]-a[1]+a[2]+a[3]);
		cnt.insert(a[0]-a[1]+a[2]-a[3]);
		cnt.insert(a[0]-a[1]+a[2]*a[3]);
		cnt.insert(a[0]-a[1]-a[2]+a[3]);
		cnt.insert(a[0]-a[1]-a[2]-a[3]);
		cnt.insert(a[0]-a[1]-a[2]*a[3]);
		cnt.insert(a[0]-a[1]*a[2]+a[3]);
		cnt.insert(a[0]-a[1]*a[2]-a[3]);
		cnt.insert(a[0]-a[1]*a[2]*a[3]);


		cnt.insert(a[0]*a[1]+a[2]+a[3]);
		cnt.insert(a[0]*a[1]+a[2]-a[3]);
		cnt.insert(a[0]*a[1]+a[2]*a[3]);
		cnt.insert(a[0]*a[1]-a[2]+a[3]);
		cnt.insert(a[0]*a[1]-a[2]-a[3]);
		cnt.insert(a[0]*a[1]-a[2]*a[3]);
		cnt.insert(a[0]*a[1]*a[2]+a[3]);
		cnt.insert(a[0]*a[1]*a[2]-a[3]);
		cnt.insert(a[0]*a[1]*a[2]*a[3]);

	}while(next_permutation(a.begin(), a.end()));
	cout << cnt.size() << "\n";
	return 0;
}