AtCoder Beginner Contest 165

传送门：https://atcoder.jp/contests/abc165/tasks

A

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

int main(){
	int k, a, b; cin>>k>>a>>b;
	for(int i=a; i<=b; i++) if(i%k==0){
		puts("OK");
		return 0;
	}	
	puts("NG");
	
	return 0;
}

B

不要做精度战士了好吗qwq

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

int n, m, q;

struct node{
	int l, r, d, v;
}e[55];

int tmp;
int a[15];
void dfs(int u, int k){
	if(u==n+1){
		int t=0;
		for(int i=1; i<=q; i++) if(a[e[i].r]-a[e[i].l]==e[i].d) t+=e[i].v;
		tmp=max(tmp, t);
		return;
	}
	
	a[u]=k;
	for(int i=k; i<=m; i++)
		dfs(u+1, i);
}

int main(){
	cin>>n>>m>>q;
	for(int i=1; i<=q; i++){
		int a, b, c, d; cin>>a>>b>>c>>d;
		e[i]={a, b, c, d};
	}
	
	int res=0;
	for(int i=1; i<=m; i++){
		tmp=0;
		dfs(1, i);
		res=max(res, tmp);
	}
	cout<<res<<endl;
	
	return 0;
}

C

注意到值域很小，直接暴力。

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

int n, m, q;

struct node{
	int l, r, d, v;
}e[55];

int tmp;
int a[15];
void dfs(int u, int k){
	if(u==n+1){
		int t=0;
		for(int i=1; i<=q; i++) if(a[e[i].r]-a[e[i].l]==e[i].d) t+=e[i].v;
		tmp=max(tmp, t);
		return;
	}
	
	a[u]=k;
	for(int i=k; i<=m; i++)
		dfs(u+1, i);
}

int main(){
	cin>>n>>m>>q;
	for(int i=1; i<=q; i++){
		int a, b, c, d; cin>>a>>b>>c>>d;
		e[i]={a, b, c, d};
	}
	
	int res=0;
	for(int i=1; i<=m; i++){
		tmp=0;
		dfs(1, i);
		res=max(res, tmp);
	}
	cout<<res<<endl;
	
	return 0;
}

D

推式子。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

#define int long long

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

int a, b, n;

signed main(){
	cin>>a>>b>>n;
	
	int r=min(n, b-1);
	cout<<a*r/b<<endl;
	
    return 0;
}

E

奇怪的构造题，奇数情况很简单，但偶数情况很难，待补。

F

直接按 dfs 序搞即可，用树状数组维护，注意到回溯的时候无法原路修改，那我们就拿个栈记录一下历史值，回溯时赋回去即可。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

#define int long long

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=2e5+5, M=N<<1;

struct node{
	int to, next;
}e[M];

int h[N], tot;

void add(int u, int v){
	e[tot].to=v, e[tot].next=h[u], h[u]=tot++;
}

int w[N], n;
int res[N];

vector<int> nums;
int a[N];

int find(int x){
	return lower_bound(all(nums), x)-nums.begin()+1;
}

int tr[N];

int lowbit(int x){return x&-x;}

int query(int p){
	int res=0;
	for(; p; p-=lowbit(p)) res=max(res, tr[p]);
	return res;
}

struct Buf{
	int p, v;
}stk[N][25];
int top, cnt[N];

void update(int p, int v){
	++top;
	for(; p<N; p+=lowbit(p)){
		++cnt[top];
		stk[top][cnt[top]].p=p;
		stk[top][cnt[top]].v=tr[p];
		tr[p]=max(tr[p], v);	
	}
}

void reset(){
	rep(i,1,cnt[top]){
		int p=stk[top][i].p;
		int v=stk[top][i].v; 
		stk[top][i]={0, 0};
		tr[p]=v;
	}
	cnt[top]=0, top--;
}

void dfs(int u, int fa){ 
	res[u]=1+query(a[u]);
	update(a[u]+1, res[u]);
	
	for(int i=h[u]; ~i; i=e[i].next){
		int go=e[i].to;
		if(go==fa) continue;
		
		dfs(go, u);
	}
	reset();
}

void work(int u, int fa){
	res[u]=max(res[u], res[fa]);
	for(int i=h[u]; ~i; i=e[i].next) if(e[i].to!=fa) work(e[i].to, u);
}

signed main(){
	memset(h, -1, sizeof h);
	cin>>n;
	rep(i,1,n) read(w[i]), nums.pb(w[i]);
	
	sort(all(nums));
	nums.erase(unique(all(nums)), nums.end());
	
	rep(i,1,n) a[i]=find(w[i]);
	
	rep(i,1,n-1){
		int u, v; read(u), read(v);
		add(u, v), add(v, u);
	}
	
	dfs(1, 0);
	work(1, 0);
	
	rep(i,1,n) cout<<res[i]<<endl;
	
    return 0;
}