ABC448 题解

A. chmin

code

#include<bits/stdc++.h>
using namespace std;
int a[1000000];
int main(){
	int N,X;
	cin >> N >> X;
	for(int i = 1; i <= N; ++i){
		cin >> a[i];
		if(a[i] < X) X = a[i], cout << "1\n";
		else cout << "0\n";
	}
}

B. Pepper Addiction

code

#include<bits/stdc++.h>
using namespace std;
const int NN = 1008;
int a[NN];
int main(){
	int n,m;
	cin >> n >> m;
	for(int i = 1; i <= m; ++i)
		cin >> a[i];
	int ans = 0;
	for(int i = 1,x,y; i <= n; ++i){
		cin >> x >> y;
		ans += min(a[x], y);
		a[x] = max(a[x] - y, 0);
	}
	cout << ans;
}

C. Except and Min

code

#include<bits/stdc++.h>
using namespace std;
const int NN = 3e5 + 8;
struct Num{
	int num,pos;
	bool operator < (const Num &x)const{
		if(num != x.num) return num < x.num;
		else return pos < x.pos;
	}
}a[NN];
int b[NN];
int n,q,k;
int main(){
	ios::sync_with_stdio(false),cin.tie(0);
	cin >> n >> q;
	for(int i = 1; i <= n; ++i){
		cin >> a[i].num;
		a[i].pos = i;
	}
	sort(a+1,a+1+n);
	while(q--){
		cin >> k;
		for(int i = 1; i <= k; ++i){
			cin >> b[i];
		}
		int ans = k + 1;
		for(int i = 1; i <= k && i <= n; ++i){
			bool vis = false;
			for(int j = 1; j <= k; ++j){
				if(a[i].pos == b[j]){
					vis = true; break;
				}
			}
			if(!vis) {ans = i;break;}
		}
		cout << a[ans].num << "\n";
	}
}

D. Integer-duplicated Path

做一遍 dfs，用 map 记录路径上经过的所有点的点权以及次数

code

#include<bits/stdc++.h>
using namespace std;
const int NN = 2e5 + 8;
int n;
int a[NN];
vector<int> e[NN];
map<int,int> mp;
bool ans[NN];
void dfs(int u,int fa){
	++mp[a[u]];
	if(mp[a[u]] > 1 || ans[fa]) ans[u] = 1;
	for(auto v : e[u]){
		if(v == fa) continue;
		dfs(v,u);
	}
	--mp[a[u]];
}
int main(){
	ios::sync_with_stdio(false),cin.tie(0);
	cin >> n;
	for(int i = 1; i <= n; ++i)
		cin >> a[i];
	for(int i = 1,u,v; i <= n-1; ++i){
		cin >> u >> v;
		e[u].push_back(v);
		e[v].push_back(u);
	}
	dfs(1,0);
	for(int i = 1; i <= n; ++i){
		if(ans[i]) cout << "Yes\n";
		else cout << "No\n";
	}
}

E. Simple Division

\[\lfloor \frac N M\rfloor \mod 10007 = \lfloor \frac {N \mod (M\times 10007)} M\rfloor \]

这样就可以使用快速幂等手段迅速计算了

code

#include<bits/stdc++.h>
using namespace std;
const int MOD = 10007;
typedef long long ll;
int k,m;
ll mod;
ll ksm(ll x, ll p){
	ll res = 1;
	while(p){
		if(p&1) res = res * x % mod;
		x = x * x % mod;
		p >>= 1;
	}
	return res;
}
ll modify(ll c,ll l){
	ll res = 0;
	ll x = 10;
	ll now = 1;
	while(l){
		if(l & 1) res = (res * x + now) % mod;
		now = now * (x+1) % mod;
		x = x * x % mod;
		l >>= 1;
	}
	return res * c % mod;
}
int main(){
	ios::sync_with_stdio(false),cin.tie(0);
	cin >> k >> m;
	mod = MOD * m;
	ll now = 0;
	for(int i = 1,c,l; i <= k; ++i){
		cin >> c >> l;
		now = now * ksm(10,l) % mod;
		now = now + modify(c,l);
		now %= mod;
//		cout << modify(c,l) << ":" << now << endl;
	}
	cout << now / m;
//	cout << mod;
}

F. Authentic Traveling Salesman Problem

tag: 构造

我们可以按列进行分块，每次一列弄完弄下一列

设列宽为B，\(T = 2\times 10^7\)

则总长度为：\(\frac {T^2} B + NB\leq 2T\sqrt N\approx 8.8\times 10^9\)

code

```#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll T = 2e7 + 8, NN = 6e4 + 8;
const ll B = T / sqrt(NN);
int n;
struct Node{
	int x,y;
	int num;
	bool operator < (const Node &A)const{
		if(x/B != A.x / B) return x/B < A.x/B;
		int block = x/B;
		return (block&1) ? y > A.y : y < A.y;
	}
}node[NN];
int main(){
	ios::sync_with_stdio(false),cin.tie(0);
	cin >> n;
	
	for(int i = 1; i <= n; ++i){
		cin >> node[i].x >> node[i].y;
		node[i].num = i;
	}
	
	sort(node+2,node+1+n);
	
	for(int i = 1; i <= n; ++i) cout << node[i].num << " ";
} 
</details>


## G. Conquest

tag: `博弈论`