AtCoder Beginner Contest 328

链接：Toyota Programming Contest 2023#7（AtCoder Beginner Contest 328） - AtCoder

A

题意：给定n个数，将小于等于x的数加起来输出。

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

void solve(){
    int n,x;
    cin>>n>>x;
    int ans=0;
    for(int i=0;i<n;i++){
        int y;
        cin>>y;
        if(y<=x) ans+=y;
    }
    cout<<ans<<endl;
}


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

B

(开始还把题意理解错了)

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

void solve(){
    int n;
    cin>>n;
    int ans=0;
    for(int i=1;i<=n;i++){
        int y;
        cin>>y;
        int z=i;
        int x=i%10;
        while(z){
            if(x!=z%10){
                x=-1;
                break;
            }
            z/=10;
        }
        if(x==-1) continue;
        z=x;
        while(z<=y){
            ans++;
            z=z*10+x;
        }
    }
    cout<<ans<<endl;
}


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

C

前缀和预处理一下就行。

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

const int N = 3e5 + 10;
int pre[N];

void solve(){
    int n,q;
    string s;
    cin>>n>>q;
    cin>>s;
    s=" " + s;
    for(int i=1;i<n;i++)
        if(s[i]==s[i+1]) pre[i]++;
    for(int i=1;i<=n;i++) pre[i]+=pre[i-1];
    for(int i=0;i<q;i++){
        int l,r;
        cin>>l>>r;
        cout<<pre[r-1]-pre[l-1]<<endl;
    }
}


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

D

思路有点向括号匹配，将字符压进栈。如果遇见ABC就把ABC弹出。

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

void solve(){
    string s;
    deque<char> path;
    cin>>s;
    for(int i=0;i<s.size();i++){
        path.push_back(s[i]);
        if(path.size()>=3){
            int len=path.size();
            if(path[len-1]=='C'&&path[len-2]=='B'&&path[len-3]=='A')
                for(int i=0;i<3;i++) path.pop_back();
        }
    }
    for(auto v:path) cout<<v;
}


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

E

在m条边里面选n-1条构成生成树。

可以深搜把所有选择都找出来然后暴力试试能不能构成生成树。

AC代码

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

const int N = 10100;
struct edg{
    int u,v,w;
}edge[N];
int n,m,k;
int fa[N];
vector<vector<int>> check;
vector<int>         path;
int Find(int x){
    return x==fa[x] ? x :fa[x]=Find(fa[x]);
}

void dfs(int starindex,int deepth){
    if(deepth==n){
        check.push_back(path);
        return;
    }
    for(int i=starindex;i<m;i++){
        path.push_back(i);
        dfs(i+1,deepth+1);
        path.pop_back();
    }
}

void solve(){
    cin>>n>>m>>k;
    for(int i=0;i<m;i++) cin>>edge[i].u>>edge[i].v>>edge[i].w;
    dfs(0,1);
    int ans=k+1;
    //cout<<check.size()<<endl;
    for(auto q:check){
        for(int i=1;i<=n;i++) fa[i]=i;
        int sum=0;
        bool OK=true;
        for(auto x:q){
            int u=edge[x].u;
            int v=edge[x].v;
            int w=edge[x].w;
            int ta=Find(u);
            int tb=Find(v);
            if(ta==tb){
                OK=false;
                break;
            }
            fa[ta]=tb;
            sum=(sum+w)%k;
        }
        if(OK)ans=min(sum,ans);
    }
    cout<<ans<<endl;
}


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

F

一个很典的带权并查集

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

const int N = 2e5 + 10;
int fa[N];
int w[N];

int Find(int x){
    if(fa[x]!=x){
        int pa = Find(fa[x]);
        w[x] += w[fa[x]];
        fa[x] = pa;
    }
    return fa[x];
}

void solve(){
    int n,q;
    cin>>n>>q;
    for(int i=1;i<=n;i++){
        fa[i]=i;
    }
    for(int i=0;i<q;i++){
        int a,b,d;
        cin>>a>>b>>d;
        int ta=Find(a);
        int tb=Find(b);
        if(ta!=tb){
            w[ta] = w[b] + d - w[a];
            fa[ta] = tb;
            cout<<i+1<<" ";
        }else if(w[a]-w[b]==d)
            cout<<i+1<<" ";
    }
}

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