winter 2024 day5

SMU 2024 winter round1

7-1最好的文档

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define int long long
 4 //#define int __int128
 5 #define double long double
 6 typedef pair<int,int>PII;
 7 typedef pair<string,int>PSI;
 8 typedef pair<string,string>PSS;
 9 const int N=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
10 const int MAXN=1e8+5;
11 const double eps=1e-12;
12 void solve() {
13     cout<<"Good code is its own best documentation.";
14 }
15 
16 signed main(){
17     ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
18     int t=1;
19 //    cin>>t;
20     while(t--){
21         solve();
22     }
23     return 0;
24 }
View Code

 

7-2自动编程

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define int long long
 4 //#define int __int128
 5 #define double long double
 6 typedef pair<int,int>PII;
 7 typedef pair<string,int>PSI;
 8 typedef pair<string,string>PSS;
 9 const int N=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
10 const int MAXN=1e8+5;
11 const double eps=1e-12;
12 void solve() {
13     int n;cin>>n;
14     cout<<"print("<<n<<")";
15 }
16 
17 signed main(){
18     ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
19     int t=1;
20 //    cin>>t;
21     while(t--){
22         solve();
23     }
24     return 0;
25 }
View Code

 

7-3程序员买包子

#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=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
void solve() {
    int n,m,k;
    string x;
    cin>>n>>x>>m>>k;
    if(k==n)cout<<"mei you mai "<<x<<" de";
    else if(k==m)cout<<"kan dao le mai "<<x<<" de";
    else cout<<"wang le zhao mai "<<x<<" de";
}

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

 

7-4猜数字-交互版

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define int long long
 4 //#define int __int128
 5 #define double long double
 6 typedef pair<int,int>PII;
 7 typedef pair<string,int>PSI;
 8 typedef pair<string,string>PSS;
 9 const int N=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
10 const int MAXN=1e8+5;
11 const double eps=1e-12;
12 void solve() {
13     int n;
14     cin>>n;
15     int l=1,r=n,x;
16     while(l<=r){
17         int mid=l+r>>1;
18         cout<<mid<<'\n';
19         string s;
20         cin>>s;
21         if(s=="<")r=mid-1;
22         else l=mid+1,x=mid;
23     }
24     cout<<"! "<<x<<'\n';
25 }
26 
27 signed main(){
28 //    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
29     int t=1;
30 //    cin>>t;
31     while(t--){
32         solve();
33     }
34     return 0;
35 }
View Code

 

7-5斯德哥尔摩火车上的题

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define int long long
 4 //#define int __int128
 5 #define double long double
 6 typedef pair<int,int>PII;
 7 typedef pair<string,int>PSI;
 8 typedef pair<string,string>PSS;
 9 const int N=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
10 const int MAXN=1e8+5;
11 const double eps=1e-12;
12 void solve() {
13     string a,b;
14     cin>>a>>b;
15     auto P=[](string a){
16         string s="";
17         for (int i = 1; i < a.size(); i++) {
18             if (a[i] % 2 == a[i-1] % 2) {
19                 s += max(a[i], a[i-1]);
20             }
21         }
22 //        s="www.multisoft.se/"+s;
23         return s;
24     };
25     a=P(a),b=P(b);
26     if(a==b)cout<<a;
27     else cout<<a<<'\n'<<b;
28 }
29 
30 signed main(){
31 //    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
32     int t=1;
33 //    cin>>t;
34     while(t--){
35         solve();
36     }
37     return 0;
38 }
View Code

 

7-6剪切粘贴

思路：模拟每一步

#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=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
void solve() {
    string s;
    cin>>s;
    int n;
    cin>>n;
    while(n--){
        int l,r;
        string a,b;
        cin>>l>>r>>a>>b;
        string c=s.substr(l-1,r-l+1);
        string now=s.substr(0,l-1)+s.substr(r,s.size()-r);
        string pp=a+b;
        int p=now.find(pp);
        if(p>=0&&p<now.size()){
            now.insert(p+a.size(),c);
        }else now+=c;
        s=now;
    }
    cout<<s;
}

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

 

7-7天梯赛的善良

#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=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
void solve() {
    int n;
    cin>>n;
    map<int,int>mp;
    for(int i=0;i<n;++i){
        int x;
        cin>>x;
        mp[x]++;
    }
    cout<<mp.begin()->first<<' '<<mp.begin()->second<<'\n';
    cout<<mp.rbegin()->first<<' '<<mp.rbegin()->second;
}

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

 

7-8谷歌的招聘

思路：枚举每个k位数判断是否为素数

#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=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
void solve() {
    int n,k;
    cin>>n>>k;
    string s;
    cin>>s;
    string a;
    auto P=[](int n){
        if(n==1||n==0)return false;
        for(int i=2;i*i<=n;++i){
            if(n%i==0)return false;
        }
        return true;
    };
    for(int i=0;i<s.size();++i){
        a.push_back(s[i]);
        if(a.size()==k){
            int b= stoi(a);
            if(P(b)){
                cout<<a;
                return ;
            }
            a.erase(a.begin());
        }
    }
    cout<<404;
}

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

 

7-9锦标赛

思路：看成满二叉树，给了每局的败者分数a，a要满足不小于左右子树的最大值（即使是该局败者，但也是其子树的胜者），找到左右子树对应的位置，有空位则插入即可。若a小于左右子树的最大值，说明无解

 

下图为每个节点对应到一维数组中的位置范围

 

#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=100+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
const int dx[4]={-1,1,0,0};
const int dy[4]={0,0,-1,1};



void solve() {
    int k;cin>>k;
    vector<int>ans(1<<k),f(1<<k);
    for(int i=k-1;i>=0;--i){
        for(int j=0;j<1<<i;++j){
            int x;
            cin>>x;
            bool ok=false;
            if(f[j*2]<=x){
                for(int p=(j*2)*(1<<k-i-1);p<(j*2+1)*(1<<k-i-1);++p){
                    if(!ans[p]){
                        ans[p]=x;
                        ok=true;
                        break;
                    }
                }
            }
            if(!ok&&f[j*2+1]<=x){
                for(int p=(j*2+1)*(1<<k-i-1);p<(2*j+2)*(1<<k-i-1);++p){
                    if(!ans[p]){
                        ans[p]=x;
                        ok=true;
                        break;
                    }
                }
            }
            if(!ok){
                cout<<"No Solution\n";
                return ;
            }
            f[j]=max({x,f[j*2],f[j*2+1]});
        }
    }
    int x;
    cin>>x;
    if(f[0]<=x){
        for(int i=0;i<1<<k;++i){
            if(!ans[i])ans[i]=x;
            cout<<ans[i];
            if(i<(1<<k)-1)cout<<' ';
        }
    }else cout<<"No Solution";
}

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

 

7-10插松枝

思路：模拟题，细心点

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define int long long
 4 //#define int __int128
 5 #define double long double
 6 typedef pair<int,int>PII;
 7 typedef pair<string,int>PSI;
 8 typedef pair<string,string>PSS;
 9 const int N=3e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
10 const int MAXN=1e8+5;
11 const double eps=1e-12;
12 void solve() {
13     int n,m,k;
14     cin>>n>>m>>k;
15     vector<vector<int>>ans;
16     stack<int>st,now;
17     vector<int>ve(n+5);
18     for(int i=0;i<n;++i)cin>>ve[i];
19     auto P=[&](){
20         vector<int>g;
21         while(now.size()){
22             g.push_back(now.top());
23             now.pop();
24         }
25         std::reverse(g.begin(), g.end());
26         ans.push_back(g);
27     };
28     for(int i=0;i<n;){
29         if(now.size()==k)P();
30         if(now.size()==0){
31             if(st.empty()){
32                 now.push(ve[i++]);
33             }
34             else {
35                 now.push(st.top()),st.pop();
36             }
37             continue;
38         }
39         if(!st.empty()&&st.top()<=now.top()){
40             now.push(st.top());st.pop();
41             continue;
42         }
43         if(ve[i]<=now.top()){
44             now.push(ve[i++]);
45             continue;
46         }else{
47             while(ve[i]>now.top()&&i<n&&st.size()<m){
48                 st.push(ve[i++]);
49             }
50             if(i<n&&ve[i]<=now.top()){
51                 now.push(ve[i++]);
52                 continue;
53             }
54         }
55         if(now.size())P();
56     }
57     while(st.size()){
58         if(now.size()==k)P();
59 
60         if(now.empty()){
61             now.push(st.top()),st.pop();
62             continue;
63         }else{
64             if(st.top()<=now.top()){
65                 now.push(st.top()),st.pop();
66                 continue;
67             }
68         }
69         if(now.size())P();
70     }
71     if(now.size())P();
72     for(int i=0;i<ans.size();++i){
73         cout<<ans[i][0];
74         for(int j=1;j<ans[i].size();++j)cout<<' '<<ans[i][j];
75         if(i<ans.size()-1)cout<<'\n';
76     }
77 }
78 
79 signed main(){
80 //    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
81     int t=1;
82 //    cin>>t;
83     while(t--){
84         solve();
85     }
86     return 0;
87 }
View Code

 

7-11拯救007

思路：数据也不大，也比较水，dfs、bfs都可以

#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=100+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
int n,d;
bool ok=false;
vector<int>st(N);
vector<PII>ve;
int dis(PII a,PII b){
    double s=(a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second);
    return s;
}
bool can(PII a,PII b,double x){
    double s=(a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second);
    if(s>x*x)return false;
    return true;
}
bool is(int u){
    if(50-ve[u].first<=d||ve[u].first+50<=d)return true;
    if(50-ve[u].second<=d||ve[u].second+50<=d)return true;
    return false;
}
void dfs(int u){
    if(is(u)){
        ok=true;
        return ;
    }
    st[u]=1;
    for(int i=0;i<n;++i){
        if(!st[i]&&can(ve[i],ve[u],d)){
            dfs(i);
        }
    }
}
void solve() {
    cin>>n>>d;
    PII o={0,0};
    for(int i=0;i<n;++i){
        int x,y;
        cin>>x>>y;
        ve.push_back({x,y});
        if(dis(ve[i],o)*1.0<=7.5*7.5)st[i]=1;
    }
    if(d*1.0>=42.5){
        cout<<"Yes";
        return ;
    }
    for(int i=0;i<n;++i){
        if(ok)break;
        if(!st[i]&&can(o,ve[i],d+7.5)){
            dfs(i);
        }
    }
    if(ok)cout<<"Yes";
    else cout<<"No";
}

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

 

7-12寻宝图

思路：dfs或bfs

 

#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=100+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
const int dx[4]={-1,1,0,0};
const int dy[4]={0,0,-1,1};


void solve() {
    int n,m;cin>>n>>m;
    vector<string>s(n+5);
    for(int i=1;i<=n;++i){
        cin>>s[i];
        s[i]=' '+s[i];
    }
    int a=0,b=0;
    for(int i=1;i<=n;++i){
        for(int j=1;j<=m;++j){
            bool ok=false;
            if(s[i][j]>='1'&&s[i][j]<='9'){
                a++;
                if(s[i][j]>'1')ok=true;
                s[i][j]='a';
                queue<PII>q;
                q.push({i,j});
                while(q.size()){
                    auto t=q.front();q.pop();
                    int x=t.first,y=t.second;
                    for(int k=0;k<4;++k){
                        int xx=x+dx[k],yy=y+dy[k];
                        if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&s[xx][yy]>='1'&&s[xx][yy]<='9'){
                            q.push({xx,yy});
                            if(s[xx][yy]>'1')ok=true;
                            s[xx][yy]='a';
                        }
                    }
                }
                if(ok)b++;
            }
        }
    }
    cout<<a<<' '<<b;
}

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

 

7-13垃圾箱分布

思路：垃圾箱数量不多，求每个垃圾箱到居民最短路

#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=1e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
const double g=9.8;
void solve() {
    cout<<"No Solution";
}

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

 

7-14非常弹的球

思路：代换下物理公式，分解下速度，vy=gt，s=vxt，E=mv2/2，vy=vsinθ，vx=vcosθ，一次起点到落地的距离x=2*s=2v2sinθvcosθ/g=v2sin2θ/g，当2θ为90度时x最大，由于v2=2E/m，每次改变v2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=1e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
const double g=9.8;
void solve() {
    double w,p,ans=0;
    cin>>w>>p;
    w/=100,p/=100;
    double E=1000;
    double x=2*E/w/g;
    while(x>eps){
        ans+=x;
        E*=(1-p);
        x=2.0*E/w/g;
    }
    cout<<fixed<<setprecision(3)<<ans;
}

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

 

7-15拯救007（升级版）

思路：求最短路径（跳的次数），注意忽略已经在岛上的点，由于有多解的情况可以先对第一跳的距离排序

 

#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=100+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const int MAXN=1e8+5;
const double eps=1e-12;
int n,d;
bool ok=false;
vector<int>path,D;
vector<PII>ve;
int dis(PII a,PII b){
    int s=(a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second);
    return s;
}
bool can(PII a,PII b,double x){
    int s=(a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second);
    if(s*1.0>x*x)return false;
    return true;
}
bool is(int u){
    if(ve[u].first-d<=-50||ve[u].first+d>=50)return true;
    if(ve[u].second-d<=-50||ve[u].second+d>=50)return true;
    return false;
}
void solve() {
    cin>>n>>d;
    PII o={0,0};
    vector<PII>g;
    path=vector<int>(n,-1);
    D=vector<int>(n,INF);
    for(int i=0;i<n;++i){
        int x,y;
        cin>>x>>y;
        ve.push_back({x,y});
        if(dis(ve[i],o)*1.0>7.5*7.5)g.push_back({dis(ve[i],o),i});
    }
    if(d*1.0>=42.5){
        cout<<1;
        return ;
    }
    queue<int>q;
    sort(g.begin(),g.end());

    for(int i=0;i<g.size();++i){
        if(can(ve[g[i].second],o,d+7.5)){
//            cout<<i<<":"<<dis(g[i],o)<<' '<<pow(7.5+d,2)<<'\n';
            q.push(g[i].second),D[g[i].second]=1;
            path[g[i].second]=-1;
        }
    }
    while(q.size()){
        auto t=q.front();q.pop();
        if(is(t)){
            stack<int>s;
            int p=t;
            while(p!=-1){
                s.push(p);
                p=path[p];
            }
            cout<<s.size()+1<<'\n';
            while(s.size()){
                int u=s.top();s.pop();
                cout<<ve[u].first<<' '<<ve[u].second;
                if(s.size())cout<<'\n';
            }
            return ;
        }
        for(int i=0;i<g.size();++i){
            if(D[g[i].second]==INF&&can(ve[t],ve[g[i].second],d)){
                D[g[i].second]=D[t]+1;
                path[g[i].second]=t;
                q.push(g[i].second);
            }
        }
    }
    cout<<0;
}

signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
//    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
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posted @ 2024-01-27 16:15  bible_w  阅读(3)  评论(0)  编辑  收藏  举报