2024牛客寒假算法基础集训营2

题目链接

A.

模拟

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

#define int long long
const int N=1e5+10;

void solve() {
    int n;cin>>n;
    while(n--){
        int a,b,c;cin>>a>>b>>c;
        int ans=0;
        if(a==150)ans+=1;
        else if(a==200)ans+=2;
        if(b==34||b==36||b==38||b==40)ans+=1;
        else if(b==45)ans+=2;
        if(c==34||c==36||c==38||c==40)ans+=1;
        else if(c==45)ans+=2;
        cout<<ans<<'\n';
    }
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int left=1;
    //cin>>left;
    while(left--){
        solve();
    }
}

B.

把防猫网的位置用小数来表示

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

#define int long long
const int N=1e5+10;

void solve() {
    int n,m,k;cin>>n>>m>>k;
    vector<vector<int>>g(n+5,vector<int>(m+5,0));
    for(int i=1,x,y;i<=k;i++){
        cin>>x>>y;
        g[x][y]=1;
    }
    int ans=0;
    map<pair<double,double>,int>mp;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(g[i][j]){
                double xx=(i-1+i)*1.0/2,yy=j*1.0;
                if(!mp[{xx,yy}]) {
                    mp[{xx, yy}] = 1;
                    ans++;
                }
                xx=(i+1+i)*1.0/2,yy=j*1.0;
                if(!mp[{xx,yy}]) {
                    mp[{xx, yy}] = 1;
                    ans++;
                }
                xx=i*1.0,yy=(j-1+j)*1.0/2;
                if(!mp[{xx,yy}]) {
                    mp[{xx, yy}] = 1;
                    ans++;
                }
                xx=i*1.0,yy=(j+1+j)*1.0/2;
                if(!mp[{xx,yy}]) {
                    mp[{xx, yy}] = 1;
                    ans++;
                }
            }
        }
    }
    cout<<ans<<'\n';
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int left=1;
    //cin>>left;
    while(left--){
        solve();
    }
}

E.

把1,2的位置分别存下来
从后往前遍历，取较小的位置，后面的全部去掉

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

#define int long long
const int N=1e5+10;

void solve() {
    int n;cin>>n;
    vector<int>pos_1,pos_2;
    for(int i=1,x;i<=n;i++){
        cin>>x;
        if(x==1)pos_1.push_back(i);
        else pos_2.push_back(i);
    }
    if(pos_2.size()==0){
        cout<<pos_1.size()<<'\n';return ;
    }
    if(pos_1.size()==0){
        cout<<pos_2.size()<<'\n';return ;
    }
    int ans=0;
    while(pos_1.size()&&pos_2.size()){
        if(pos_1.back()<pos_2.back()){
            ans++;
            while(pos_2.size()&&pos_1.back()<pos_2.back())pos_2.pop_back();
            pos_1.pop_back();
        }
        if(pos_2.back()<pos_1.back()){
            ans++;
            while(pos_1.size()&&pos_2.back()<pos_1.back())pos_1.pop_back();
            pos_2.pop_back();
        }
    }
    if(pos_2.size())ans+=pos_2.size();
    if(pos_1.size())ans+=pos_1.size();
    cout<<ans<<'\n';
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int left=1;
    cin>>left;
    while(left--){
        solve();
    }
}

I.

可以发现边权的计算方式就是两倍的较大值

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

#define int long long
const int N=1e5+10;

void solve() {
    int n;cin>>n;
    vector<int>a(n+1);
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    sort(a.begin()+1,a.end());
    int ans=0;
    for(int i=2;i<=n;i++){
        ans+=(i-1)*2*a[i]*2;
    }
    cout<<ans<<'\n';
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int left=1;
    cin>>left;
    while(left--){
        solve();
    }
}

J.

可以发现边权计算方式就是两倍的较小值或最小值的4倍

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

#define int long long
const int N=1e5+10;

void solve() {
    int n;cin>>n;
    vector<int>v(n+1);
    int mi=0x3f3f3f3f;
    for(int i=1;i<=n;i++){
        cin>>v[i];
    }
    sort(v.begin()+1,v.end());
    mi=v[1];
    for(int i=1;i<=n;i++){
        if(v[i]==mi)v[i]*=2;
        else v[i]=min(2*v[i],4*mi);
    }
    vector<int>pre(n+1,0);
    for(int i=1;i<=n;i++)pre[i]=pre[i-1]+v[i];
    int ans=0;
    for(int i=2;i<=n;i++){
        ans+=pre[i-1];
    }
    ans*=2;
    cout<<ans<<'\n';
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int left=1;
    cin>>left;
    while(left--){
        solve();
    }
}