我怎么又是这个点写题解

这套题还是挺简单的（确信

A

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

int main(){
	int res;
	for(int i=1; i<=5; i++){
		int t; cin>>t;
		if(!t) res=i;
	}
	cout<<res;
	return 0;
}

B

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

int main(){
	int x, y; cin>>x>>y;
	if(y&1){
		puts("No");
		return 0;
	}
	
	puts(y/2<=2*x && 2*x<=y? "Yes": "No");
	
	return 0;
}

C

边界注意一下嗷。

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

const int N=105;

int buc[N];

int main(){
	int x, n; cin>>x>>n;
	for(int i=1; i<=n; i++){
		int t; cin>>t;
		buc[t]++; 
	}
	
	int res=1e9, ans;
	for(int i=0; i<=101; i++) if(!buc[i] && abs(i-x)<res) res=abs(i-x), ans=i; 
	cout<<ans;
	
	return 0;
}

D

排序，大力去筛就行了，复杂度是 \(O(NlogN)\) ，但是要记得判重。

#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;

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=5*N;

int n;
int w[N], vis[M], cnt[M];

int main(){
	read(n);
	rep(i,1,n) read(w[i]), cnt[w[i]]++;
   
    sort(w+1, w+1+n);
   
    int res=0;
    rep(i,1,n){
    	int v=w[i];
    	if(vis[v]) continue;
    	if(!vis[v] && cnt[v]==1) res++;
    	
    	for(int j=v; j<M; j+=v) vis[j]=true;
    }
    
    cout<<res;
    
   
	return 0; 
}

E

写了个树套树维护。

#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;

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, cnt=2e5;

int n, q;

struct node{
	int l, r;
	int v;
	multiset<int> val;
	
	#define ls(u) u<<1
	#define rs(u) u<<1|1
}tr[N<<2];

int pos[N], rat[N]; // msg of kid

void pushup(int u){
	tr[u].v=min(tr[ls(u)].v, tr[rs(u)].v);
}

void build(int u, int l, int r){
	tr[u]={l, r, INF};
	if(l==r) return;
	int mid=l+r>>1;
	build(ls(u), l, mid), build(rs(u), mid+1, r);
	pushup(u);
}

void modify(int u, int x, bool fl, int v){
	if(tr[u].l==tr[u].r && tr[u].l==x){
		if(fl){
			auto &t=tr[u].val;
			t.insert(v);
			tr[u].v=*t.rbegin();
		}
		else{
			auto &t=tr[u].val;
			auto it=t.find(v);
			t.erase(it);
			
			if(t.size()) tr[u].v=*t.rbegin();
			else tr[u].v=INF;
		}
		return;
	}
	
	int mid=tr[u].l+tr[u].r>>1;
	if(mid>=x) modify(ls(u), x, fl, v);
	else modify(rs(u), x, fl, v);
	pushup(u);
}

int query(int u, int l, int r){
	if(tr[u].l>=l && tr[u].r<=r) return tr[u].v;
	
	int mid=tr[u].l+tr[u].r>>1, res=INF;
	if(mid>=l) res=min(res, query(ls(u), l, r));
	if(mid<r) res=min(res, query(rs(u), l, r));
	return res;
}

int main(){
	read(n), read(q);
	
	build(1, 1, cnt);
	
	rep(i,1,n){
		int v, p; read(v), read(p);
		modify(1, p, 1, v); // insert
		pos[i]=p, rat[i]=v;
	}
	
	rep(i,1,q){
		int idx, p; read(idx), read(p);
		int pre=pos[idx], rate=rat[idx]; // pre pos, rate
		
		modify(1, pre, 0, rate); // delete
		pos[idx]=p;
		modify(1, p, 1, rate); // insert
		
		int res=query(1, 1, cnt);
		cout<<res<<endl;
	}
	
    return 0;
}

F

最短路。

#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;

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=1e6+5;

int n, m, k;

int sx, sy, tx, ty;
string g[N];

PII q[N];
int tt=-1, hh=0;

int dx[]={1, 0, -1, 0}, dy[]={0, 1, 0, -1};

int main(){
	cin>>n>>m>>k>>sx>>sy>>tx>>ty;
	sx--, sy--, tx--, ty--;
	
	vector<vector<int>> d(n, vector<int>(m, INF));
	vector<vector<char>> g(n, vector<char>(m));
	
	getchar();
	for(int i=0; i<n; i++, getchar()) for(int j=0; j<m; j++) g[i][j]=getchar();
	
	d[sx][sy]=0;
	q[++tt]={sx, sy};
	
	// rep(i,0,n-1) rep(j,0,m-1) debug(g[i][j]);
	
	while(tt>=hh){
		auto hd=q[hh++];
		int x=hd.first, y=hd.second; 
		rep(i,0,3){
			rep(j,1,k){
				int kx=x+dx[i]*j, ky=y+dy[i]*j;
				
				if(kx<0 || kx>=n || ky<0 || ky>=m || d[kx][ky]<=d[x][y]) break;
				if(g[kx][ky]==‘@‘) break;
			
				if(d[kx][ky]==INF) q[++tt]={kx, ky}, d[kx][ky]=d[x][y]+1;
				if(kx==tx && ky==ty) break;
			}
		}
	}
	
	cout<<(d[tx][ty]==INF? -1: d[tx][ty])<<endl;
	
    return 0;
}

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