23 暑假友谊赛 No.4

23 暑假友谊赛 No.4

A - Alien Sunset

思路：用差分数组标记每个周期的夜晚，（从0开始）

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;
const int m=1825*100;
void solve(){
    int n;cin>>n;
    vector<int>cnt(m+110,0);
    for(int i=1,l,r,h;i<=n;++i){
        cin>>h>>l>>r;
        if(l<=r){
            int s=0,t=h-1;
            while(s<=m){
                cnt[s]++,cnt[l+1]--;
                cnt[r]++,cnt[t+1]--;
                s+=h,t+=h,l+=h,r+=h;
            }
        }else{
            while(r<=m){
                cnt[r]++,cnt[l+1]--;
                r+=h,l+=h;
            }
        }
    }
    for(int i=1;i<m;++i)cnt[i]+=cnt[i-1];
    for(int i=0;i<m;++i){
        if(cnt[i]==n){
            cout<<i;return ;
        }
    }
    cout<<"impossible";
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

 

B - Breaking Biscuits

 思路：枚举每条边，以该边为平行杯壁的边，将所有的点以该边为界线分为两部分，分别找到两部分中最远的点，两点与边的距离和即为当前情况的最小杯直径，取所有情况的直径的最小值

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=100+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353,M=44750;
const int inf=0x3f3f3f3f3f3f;
const double eps=1e-6;

struct E{
    int x,y;
    E(){}
    E(int _x,int _y){x=_x,y=_y;}
    E operator-(E b){return E(x-b.x,y-b.y);}
    double len(){return ::hypot(x,y);}
};
void solve(){
    int n;
    cin>>n;
    double ans=INF;
    vector<E>ve(n);
    for(int i=0;i<n;++i)cin>>ve[i].x>>ve[i].y;
    auto cross=[](E a,E b){
        return (double)(a.x*b.y-a.y*b.x);
    };
    auto dis=[&](E p,E a,E b){
        return cross(p-a,b-a)/(b-a).len();
    };
    for(int i=0;i<n;++i)
        for(int j=0;j<i;++j){
            double l=INF,r=-INF;
            for(int k=0;k<n;++k){
                l=min(l,dis(ve[k],ve[i],ve[j]));
                r=max(r,dis(ve[k],ve[i],ve[j]));
            }
            r-=l;
            ans=min(ans,r);
        }
    cout<<fixed<<setprecision(10)<<ans;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}

 

C - Cued In

思路：每次打red后打的球可再返回桌子，求出red的个数s，非red的最大分值ma，除red外的所有分值le，那么答案为s（1+ma）+le

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;
const int m=1825*100;
void solve(){
    map<string,int>mp;
    mp["red"]=1,mp["yellow"]=2,mp["green"]=3,mp["brown"]=4,mp["blue"]=5,mp["pink"]=6,mp["black"]=7;
    int n,ma=0,cnt=0,le=0;cin>>n;
    string s;
    for(int i=0;i<n;++i){
        cin>>s;
        ma=max(ma,mp[s]);
        if(s=="red")cnt++;
        else le+=mp[s];
    }
    if(le)cout<<cnt*(1+ma)+le;
    else cout<<1;
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

 

D - Deranging Hat

思路：给了t，s为t排序后的串，问s转为t的操作，维护s和t的前i个字符都相同，若不相同则往后找到与之交换的字符即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;
const int m=1825*100;
void solve(){
    string t;cin>>t;
    string s=t;
    sort(s.begin(),s.end());
    for(int i=0;i<s.size();++i){
        if(s[i]!=t[i]){
            for(int j=i+1;j<s.size();++j){
                if(s[j]==t[i]){
                    if(s[i]>=s[j])cout<<i+1<<' '<<j+1<<'\n';
                    else cout<<j+1<<' '<<i+1<<'\n';
                    swap(s[i],s[j]);
                    break;
                }
            }
        }
    }
    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 - Education

思路：将房子按价格少、容量小排序，将部门容量按大到小排序，暴力枚举即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,m;cin>>n>>m;
    vector<PII>ve(n);
    vector<pair<PII,int>>g(m);
    for(int i=0;i<n;++i)cin>>ve[i].first,ve[i].second=i+1;
    for(int i=0;i<m;++i)cin>>g[i].first.second;
    for(int i=0;i<m;++i)cin>>g[i].first.first,g[i].second=i+1;
    sort(ve.begin(),ve.end(),greater<PII>());
    sort(g.begin(),g.end());
    vector<int>st(m+1,0),ans(n+1);
    for(int i=0;i<n;++i){
        for(int j=0;j<m;++j){
            if(!st[g[j].second]&&g[j].first.second>=ve[i].first){
                ans[ve[i].second]=g[j].second;st[g[j].second]=1;
                break;
            }
            if(j==m-1){cout<<"impossible";return ;}
        }
    }
    for(int i=1;i<=n;++i)cout<<ans[i]<<' ';
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

 

F - Flipping Coins

思路：求k次操作后有正面硬币的期望值，f[i][j]表示扔第i次有j枚硬币的期望值，i从0开始，即：f[i+1][j]+=f[i][j]*0.5，f[i+1][j+1]+=f[i][j]*0.5（分别为扔反面、正面）

当已有n枚正面时，f[i+1][j]+=f[i][j]*0.5，f[i+1][j-1]+=f[i][j]*0.5（分别为扔正面、反面）

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,k;cin>>n>>k;
    vector<vector<double>>f(k+1,vector<double>(n+1,0));
    f[0][0]=1;
    for(int i=0;i<k;++i)
        for(int j=0;j<=min(n,i);++j){
            f[i+1][j]+=f[i][j]*0.5;
            if(j==n)f[i+1][j-1]+=f[i][j]*0.5;
            else f[i+1][j+1]+=f[i][j]*0.5;
        }
    double ans=0;
    for(int i=1;i<=n;++i)ans+=f[k][i]*i;
    cout<<fixed<<setprecision(10)<<ans;
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

 

G - Gentlebots

思路：靠近终点依次走x，y，z方向，若走了一步后发现与另一个机器重合，返回后走另一个方向即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;
struct E{
    int x,y,z;
}s1,s2,t1,t2;
bool eq(E a,E b){
    return a.x==b.x&&a.y==b.y&&a.z==b.z;
}
void pr(E s,E t){
    cout<<"("<<s.x<<' '<<s.y<<' '<<s.z<<") ";
    cout<<"("<<t.x<<' '<<t.y<<' '<<t.z<<")\n";
}
void solve(){
    cin>>s1.x>>s1.y>>s1.z>>t1.x>>t1.y>>t1.z;
    cin>>s2.x>>s2.y>>s2.z>>t2.x>>t2.y>>t2.z;
    pr(s1,s2);
    while(1){
        if(s1.x==t1.x&&s1.y==t1.y&&s1.z==t1.z&&s2.x==t2.x&&s2.y==t2.y&&s2.z==t2.z)break;
        E v=s1,p;p={0,0,0};
        if(v.x!=t1.x){
            if(t1.x>v.x)v.x++;else v.x--;
            if(eq(v,s2))p={0,-1,0};
        }else if(v.y!=t1.y){
            if(t1.y>v.y)v.y++;else v.y--;
            if(eq(v,s2))p={0,0,-1};
        }else if(v.z!=t1.z){
            if(t1.z>v.z)v.z++;else v.z--;
            if(eq(v,s2))p={-1,0,0};
        }
        if(!eq(p,{0,0,0}))v=s1,v.x+=p.x,v.y+=p.y,v.z+=p.z;
        s1=v;
        v=s2,p={0,0,0};
        if(v.x!=t2.x){
            if(t2.x>v.x)v.x++;else v.x--;
            if(eq(v,s1))p={0,-1,0};
        }else if(v.y!=t2.y){
            if(t2.y>v.y)v.y++;else v.y--;
            if(eq(v,s1))p={0,0,-1};
        }else if(v.z!=t2.z){
            if(t2.z>v.z)v.z++;else v.z--;
            if(eq(v,s1))p={-1,0,0};
        }
        if(!eq(p,{0,0,0}))v=s2,v.x+=p.x,v.y+=p.y,v.z+=p.z;
        s2=v;
        pr(s1,s2);
    }
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>t;
    while(T--){
        solve();
    }
    return 0;
}

 

H - Hiking

I - I Work All Day

思路：找到取模后最小的数

void solve() {
    int n; cin >> n;
    vector<int> ve(n);
    for (int i = 0; i < n; ++i) {
        cin >> ve[i];
    }
    int x; cin >> x;
    int ans = inf, res;
    for (int i = 0; i < n; ++i) {
        if( x % ve[i]  <ans) ans = x%ve[i], res = ve[i];
    }
    cout << res << '\n';
}

 

J - Just A Minim

思路：算总分

void solve() {
    int n; cin >> n;
    map<int, double> mp;
    mp[0] = 2.0, mp[1] = 1.0, mp[2] = 0.5, mp[4] = 0.25, mp[8] = 0.125, mp[16] = 0.0625;
    double ans = 0;

    for (int i = 0; i < n; ++i) {
        int x; cin >> x;
        ans += mp[x];
    }
    printf("%.8f", ans);
}

 

K - Knightsbridge Rises

L - Lizard Lounge

思路：将每个点按向量划分，按与电视的距离排序，求最长上升子序列的个数即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    PII T,v,u;
    cin>>T.first>>T.second;
    int n;cin>>n;
    map<PII,vector<PII>>mp;
    for(int i=1,h,d;i<=n;++i){
        cin>>v.first>>v.second>>h;
        v.first-=T.first,v.second-=T.second;
        d=abs(__gcd(v.first,v.second));
        u.first=v.first/d,u.second=v.second/d;
        mp[u].push_back({abs(v.first)+abs(v.second),h});
    }
    int ans=0;
    for(auto [a,b]:mp){
        sort(b.begin(),b.end());
        vector<int>q;
        q.push_back(b.begin()->second);
        for(int i=1;i<b.size();++i){
            if(b[i].second>q.back())q.push_back(b[i].second);
            else *lower_bound(q.begin(),q.end(),b[i].second)=b[i].second;
        }
        ans+=q.size();
    }
    cout<<ans;
    return;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}