AtCoder Beginner Contest 310

（AtCoder Beginner Contest 310）

 A - Order Something Else

思路：比较下打折和不打折的情况

#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int>PII;

typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e5+5,INF=0x3f3f3f3f,Mod=1e9+7;
const double eps=1e-6;

void solve(){
    int n,p,q;cin>>n>>p>>q;
    int ans=p;
    for(int i=1;i<=n;++i){
        int x;cin>>x;
        ans=min(ans,x+q);
    }
    cout<<ans;
}
signed main(){

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

B - Strictly Superior

思路：找出满足条件的任务（i，j）即YES，满足：

p[j]<=p[i]，j任务的功能需包含i所有的功能，若p[j]=p[i]，j任务还需比i任务多至少1个功能

#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int>PII;

typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e5+5,INF=0x3f3f3f3f,Mod=1e9+7;
const double eps=1e-6;

void solve(){
    int n,m;cin>>n>>m;
    set<int>se[n];
    vector<int>p(n),c(n);
    for(int i=0;i<n;++i){
        cin>>p[i]>>c[i];
        for(int j=0,x;j<c[i];++j){
            cin>>x;
            se[i].insert(x);
        }
    }
    bool ok=false;
    for(int i=0;i<n;++i){
        for(int j=0;j<n;++j){
            if(i!=j&&p[j]<=p[i]){
                int same=0;
                for(auto v:se[i]){
                    if(se[j].count(v)){
                        same++;
                    }
                }
                if(p[j]<p[i]&&same==se[i].size()){
                    ok=true;break;
                }else if(p[j]==p[i]&&same==se[i].size()&&se[j].size()>same){
                    ok=true;break;
                }
            }
        }
        if(ok)break;
    }
    if(ok)cout<<"Yes";
    else cout<<"No";
}
signed main(){

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

 C - Reversible

思路：用map标记原始有的所有字符串，询问字符串s是否含有时，看s和s的倒序后的字符串是否被标记即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int>PII;

typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e5+5,INF=0x3f3f3f3f,Mod=1e9+7;
const double eps=1e-6;

void solve(){
    int n,ans=0;cin>>n;
    string s;
    map<string,int>mp;
    for(int i=0;i<n;++i){
        cin>>s;
        string a=s;
        reverse(s.begin(), s.end());
        if(!mp[s]&&!mp[a])ans++;
        mp[a]++;
    }
    cout<<ans;
}
signed main(){

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

D - Peaceful Teams

思路：搜索所有分队可能，求所有满足条件的情况

#include<bits/stdc++.h>
using namespace std;
#define  int long long
typedef pair<int,int>PII;

typedef pair<string,int>PSI;
typedef pair<char,int>PCI;
typedef pair<string,string>PSS;
const int N=1e5+5,INF=0x3f3f3f3f,Mod=1e9+7;
const double eps=1e-6;
typedef long long ll;


vector<vector<int>>ve;
int g[11][11],res;
int n,t,m;
void dfs(int u,int cnt){
    if(cnt>t)return ;
    if(u>n){
        if(cnt==t){
            for(auto i:ve)
                for(auto a:i)
                    for(auto b:i)
                        if(g[a][b])return ;
            res++;
        }
        return ;
    }
    for(int i=1;i<=cnt+1;++i){
        ve[i].push_back(u);
        dfs(u+1,max(i,cnt));
        ve[i].pop_back();
    }
}
void solve(){
    cin>>n>>t>>m;
    ve=vector<vector<int>>(n+1,vector<int>(n+1));
    for(int i=0,a,b;i<m;++i){
        cin>>a>>b;g[a][b]=g[b][a]=1;
    }
    dfs(1,0);
    cout<<res;
    return ;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}

E - NAND repeatedly

思路：求所有区间的价值和。可以看出 a⊼b =！（a&b）。

用f（i）表示以i结尾的所有区间的价值和，将所有f（i）加起来即可

f（i）的状态转移：

若a[i]为0，f（i）=f（i-1）+（i-1）；

若a[i]为1，f（i）=（i-1）-f（i-1）；

#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int>PII;

typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e5+5,INF=0x3f3f3f3f,Mod=1e9+7;
const double eps=1e-6;

void solve(){
    int n;cin>>n;
    string s;cin>>s;
    s.insert(s.begin(),' ');
    vector<int>f(s.size(),0);
    int ans=0;
    //f[1]=!(s[1]&&s[1]);
    for(int i=1;i<=n;++i){
        if(s[i]=='0'){
            f[i]+=(i-1);
        }else{
            f[i]+=(i-1-f[i-1]);
        }
        f[i]+=(s[i]-'0');//cout<<f[i]<<' ';
        ans+=f[i];
    }
    cout<<ans;
}
signed main(){

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