winter 2024 day1

2024 蓝桥杯模拟赛 1 (div1)

 

 =.= ^-^

 

A[蓝桥杯 2021 国 BC] 大写

思路：小写转换大写

#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 double eps=1e-6;


void solve(){
    string s;
    cin>>s;
    for(int i=0;i<s.size();++i){
        if(s[i]>='a'&&s[i]<='z')s[i]-=32;
        cout<<s[i];
    }
}

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

 

 B[传智杯 #4 决赛] 小智的疑惑

思路：暴力判断

#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 double eps=1e-6;


void solve(){
    string s;
    string f="chuanzhi";
    cin>>s;
    int ans=0;
    for(int i=0;i+7<s.size();++i){
        if(s[i]=='c'){
//            cout<<i<<':'<<s[i]<<'\n';
            bool ok=true;
            for(int j=i,x=0;x<f.size();++x,++j){
                if(f[x]!=s[j]){
                    ok=false;
                    break;
                }
            }
            if(ok)ans++,i+=7;
        }
    }
    cout<<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[蓝桥杯 2017 省 B] 日期问题

思路：枚举所有情况勒，需判断日期是否合法，且日期不重复

#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 double eps=1e-6;


void solve(){
    map<string,string>mp;
    mp["01"]="31",mp["02"]="29",mp["03"]="31",mp["04"]="30",mp["05"]="31";
    mp["06"]="30",mp["07"]="31",mp["08"]="31",mp["09"]="30",mp["10"]="31";
    mp["11"]="30",mp["12"]="31";
    string s;
    cin>>s;
    string a=s.substr(0,2),b=s.substr(3,2),c=s.substr(6,2);
    string aa,bb,cc;
    vector<string>ss(3);
    if(a<="59")aa="20"+a;
    else aa="19"+a;
    bb=b,cc=c;
    ss[0]=aa+"-"+bb+"-"+cc;

    if(c<="59")cc="20"+c;
    else cc="19"+c;
    aa=a,bb=b;
    ss[1]=cc+"-"+aa+"-"+bb;
    ss[2]=cc+"-"+bb+"-"+aa;

    sort(ss.begin(),ss.end());
    map<string,string>st;
    for(int i=0;i<3;++i){
        a=ss[i].substr(5,2),b=ss[i].substr(8,2);
        c=ss[i].substr(0,4);
        if(!(a>="01"&&a<="12"))continue;
        int y=stoi(c);
        if((y%4==0&&y%100!=0)||y%400==0)mp["02"]="29";
        else mp["02"]="28";
        if(b>mp[a]||b=="00")continue;
        if(st.count(ss[i]))continue;
        cout<<ss[i]<<'\n';
        st[ss[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;
}

 

D[蓝桥杯 2023 省 B] 冶炼金属

思路：转换率的最大值为a/b的最小值，最小值为a/（b+1）+1的最大值

#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 double eps=1e-6;


void solve(){
    int n;cin>>n;
    vector<PII>ve(n);
    for(int i=0;i<n;++i)cin>>ve[i].first>>ve[i].second;
    int ma=1e9,mi=0;
    for(int i=0;i<n;++i){
        ma=min(ma,ve[i].first/ve[i].second);
        mi=max(mi,ve[i].first/(ve[i].second+1)+1);
    }
    cout<<mi<<' '<<ma;
}

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

 

E[蓝桥杯 2015 省 A] 饮料换购

思路：记录每次换瓶子后的瓶盖数，统计换得的总瓶子数即可

#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=100+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    int n;cin>>n;
    int ans=n,l=n,c;
    while(l>=3){
        c=l/3;
        l-=2*c;
        ans+=c;
    }
    cout<<ans;
}

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

 

F[蓝桥杯 2017 省 AB] 分巧克力

思路：二分边长a，一个矩形中正方形的个数为（h/a）*（w/a）

#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 double eps=1e-6;


void solve(){
    int n,k;cin>>n>>k;
    vector<int>h(n),w(n);
    int ma=0;
    for(int i=0;i<n;++i){
        cin>>h[i]>>w[i];
        ma=max(ma,min(h[i],w[i]));
    }
    int l=1,r=ma,x;
    auto check=[&](int a){
        int c=0;
        for(int i=0;i<n;++i){
            c+=(h[i]/a)*(w[i]/a);
        }
        return c>=k;
    };
    while(l<=r){
        int mid=l+r>>1;
        if(check(mid))l=mid+1,x=mid;
        else r=mid-1;
    }
    cout<<x;
}

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

 

G[蓝桥杯 2016 省 B] 交换瓶子

思路：预处理每个数对应的位置，若该位置上的数不是本身，找到目标数所在的位置进行交换，记录交换次数。

也可以用图解决...，参考https://blog.csdn.net/qq_29992017/article/details/129885881

#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=1e4+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    int n;
    cin>>n;
    vector<int>ve(n+5),s(n+5);
    int ans=0;
    for(int i=1;i<=n;++i){
        cin>>ve[i];
        s[ve[i]]=i;
    }
    for(int i=1;i<=n;++i){
        if(i==ve[i])continue;
        int b=s[i];
        s[i]=i,s[ve[i]]=b;
        swap(ve[i],ve[b]);
        ans++;
    }
    cout<<ans;
}

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

 

H[蓝桥杯 2015 省 B] 生命之树

思路：树形dp一下子

#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=1e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

int n,w[N],ans,st[N],f[N];
vector<vector<int>>g;
int dfs(int u){
    for(auto v:g[u]){
        if(st[v])continue;
        st[v]=1;
        f[v]=w[v]+dfs(v);
        if(f[v]>0)f[u]+=f[v];
    }
    return f[u];
}
void solve(){
    cin>>n;
    g=vector<vector<int>>(n+1);
    for(int i=1;i<=n;++i){
        cin>>w[i];
    }
    for(int i=1;i<n;++i){
        int u,v;
        cin>>u>>v;
        g[u].push_back(v),g[v].push_back(u);
    }
    st[1]=1;
    f[1]=w[1]+dfs(1);
    for(int i=1;i<=n;++i)ans=max(ans,f[i]);
    cout<<ans;
}

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

 

I[蓝桥杯 2016 省 A] 密码脱落

思路：可以在任意位置添加，最坏的情况是每个字符都添加一个，如何尽可能的少添加，可以将原序列与置换后的序列求最长公共子序列，这些字符就不需要添加，答案即为n-lcs（最长公共子序列）

 

#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 double eps=1e-6;

void solve(){
    string s,f;
    cin>>s;
    int n=s.size();
    f=s, std::reverse(s.begin(), s.end());
    s=' '+s;
    f=' '+f;
    vector<vector<int>>ve(n+5,vector<int>(n+5));
    for(int i=1;i<=n;++i){
        for(int j=1;j<=n;++j){
            ve[i][j]=max(ve[i-1][j],ve[i][j-1]);
            if(s[i]==f[j])ve[i][j]=max(ve[i][j],ve[i-1][j-1]+1);
        }
    }
    cout<<n-ve[n][n];
}

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

 

J[蓝桥杯 2022 省 A] 爬树的甲壳虫

思路：f[i]表示到达i的时间期望，那么f[i]=（1-p[i]）*f[i-1]+1+p[i]*（f[i-1]+f[i]），解方程得f[i]=（f[i-1]+1）/（1-p[i]）

#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 double eps=1e-6;

int ksm(int a,int b){
    int ans=1;
    while(b){
        if(b&1)ans=ans*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}
void solve(){
    int n;cin>>n;
    vector<int>p(n+1);
    for(int i=1;i<=n;++i){
        int x,y;
        cin>>x>>y;
        p[i]=y*ksm(y-x,mod-2)%mod;
    }
    vector<int>f(n+1);
    for(int i=1;i<=n;++i){
        f[i]=(f[i-1]+1)*p[i]%mod;
    }
    cout<<f[n];
}

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