Codeforces Round #799 (Div. 4)

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#c++ #算法 #数据结构

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本文探讨了五种编程问题：马拉松算法的计数，元素唯一性的处理，寻找象棋棋子位置，时钟操作模拟，以及二进制双端队列操作。通过实例展示了如何解决这些与数据结构、算法和逻辑思维相关的问题。

文章目录

A. Marathon
B. All Distinct
C. Where's the Bishop?
D. The Clock
E. Binary Deque
F. 3SUM
G. 2^Sort
H 不会

A. Marathon

在这里插入图片描述
打卡题

#include <bits/stdc++.h>
 
#define pt2(x,y) cout <<(x)<<"--"<<(y)<<endl;
#define pt1(x) cout <<"#"<<(x)<<"#"<<endl;
#define hh cout << "\n"
#define rep(i,a,n) for(int i = (a); i <= int(n); i++) 
#define per(i,a,n) for(int i = (n); i >= int(a); i--) 
#define fi first;
#define se second;
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef double DB;
const int N=1e6+10,M=1e6+10,INF=0x3f3f3f3f;
 
int t=1;
 
void solve()
{
    int a,b,c,d;
    cin >> a >> b >> c >> d;
    int ans=0;
    if(b>a) ans++;
    if(c>a) ans++;
    if(d>a) ans++;
    cout << ans <<endl;
}   
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
 
    cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

B. All Distinct

在这里插入图片描述
分奇数偶数，奇数可以自己解决最后只剩一个，偶数个数2个配对消，向下取整

#include <bits/stdc++.h>
 
#define pt2(x,y) cout <<(x)<<"--"<<(y)<<endl;
#define pt1(x) cout <<"#"<<(x)<<"#"<<endl;
#define hh cout << "\n"
#define rep(i,a,n) for(int i = (a); i <= int(n); i++) 
#define per(i,a,n) for(int i = (n); i >= int(a); i--) 
#define fi first;
#define se second;
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef double DB;
const int N=1e6+10,M=1e6+10,INF=0x3f3f3f3f;
 
int t=1;
unordered_map<int,int> h;
int cnt;
int find(int x)
{
    if(h.count(x)) return h[x];
    return h[x]=cnt++;
 
}
void solve()
{
    h.clear();
    cnt=1;
    int n;
    int st[100]={0};
    cin >> n;
    int ans=0;
    rep(i,1,n)
    {
        int x;
        cin >> x;
        st[find(x)]++;
    }
    int ji=0,ou=0;
    rep(i,1,50)
    {
        if(st[i]==0) continue;
        if(st[i]%2==0) ou++;
        else ji++; 
    }
    ans=ji+(ou/2)*2;
    cout << ans<<endl;
 
}   
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
 
    cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

C. Where’s the Bishop?

在这里插入图片描述
找前一行有2个当前这行只有一个的位置

#include <bits/stdc++.h>
 
#define pt2(x,y) cout <<(x)<<"--"<<(y)<<endl;
#define pt1(x) cout <<"#"<<(x)<<"#"<<endl;
#define hh cout << "\n"
#define rep(i,a,n) for(int i = (a); i <= int(n); i++) 
#define per(i,a,n) for(int i = (n); i >= int(a); i--) 
#define fi first;
#define se second;
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef double DB;
const int N=1e6+10,M=1e6+10,INF=0x3f3f3f3f;
 
int t=1;
char g[10][10];
void solve()
{
    rep(i,0,7) cin >> g[i];
    int last=0;
     rep(i,0,7)  
     {
        int res=0;
        int x,y;
        rep(j,0,7) 
        {
            if(g[i][j]=='#') 
            {
                res++;
                x=i,y=j;
            }
        }
        if(res==1&& last==2)  
        {
            cout << x+1<<" "<<y+1<<endl;
            break;
        }
        last=res;
     } 
 
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
 
    cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

D. The Clock

在这里插入图片描述
简单的模拟题

#include <bits/stdc++.h>
 
#define pt2(x,y) cout <<(x)<<"--"<<(y)<<endl;
#define pt1(x) cout <<"#"<<(x)<<"#"<<endl;
#define hh cout << "\n"
#define rep(i,a,n) for(int i = (a); i <= int(n); i++) 
#define per(i,a,n) for(int i = (n); i >= int(a); i--) 
#define fi first;
#define se second;
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef double DB;
const int N=1e6+10,M=1e6+10,INF=0x3f3f3f3f;
 
int t=1;
 
bool check(int i,int j)
{
    int i1=i%10,i2=i/10;
    int j1=j%10,j2=j/10;
    if(j2==i1 && j1 == i2) return true;
    return false;
}
void change(int &h,int &m,int x)
{
    int dh = x/60;
    int dm = x%60;
    h=dh+h;
    m=dm+m;
    if(m>=60) 
    {
        m=m%60;
        h++;
    }
    if(h>=24)
    {
        h=h%24;
    }
}
void solve()
{
    int h,m,x;
    scanf("%d:%d",&h,&m);
    scanf("%d",&x);
    int ans=0;
 
    int ch=h,cm=m;
    bool one=true;
    while(one||h!=ch || m !=cm)
    {
        one=false;
        change(h,m,x);
        if(check(h,m)) ans++;
    }
    cout << ans<<endl;
}   
int main()
{
    // cin.tie(0);
    // cout.tie(0);
    // ios::sync_with_stdio(0);
 
    cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

E. Binary Deque

在这里插入图片描述
前缀和加枚举区间

#include <bits/stdc++.h>
 
#define pt2(x,y) cout <<(x)<<"--"<<(y)<<endl;
#define pt1(x) cout <<"#"<<(x)<<"#"<<endl;
#define hh cout << "\n"
#define rep(i,a,n) for(int i = (a); i <= int(n); i++) 
#define per(i,a,n) for(int i = (n); i >= int(a); i--) 
#define fi first;
#define se second;
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef double DB;
const int N=1e6+10,M=1e6+10,INF=0x3f3f3f3f;
 
int t=1;
int a[N],b[N];
void solve()
{
    int n,m;
    cin >> n >> m;
    rep(i,1,n)
    {
        int x;
        cin >> x;
        a[i]=a[i-1]+x;
    }
    if(a[n]<m) 
    {
        cout <<-1<<endl;
        return;
    } 
    rep(i,1,n) b[i]=-1; // b表示值为i的数第一次出现的位置
    b[0]=0;
    int ans=n;
    rep(r,1,n) 
    {
        if(a[r]-m >= 0 && b[a[r]-m]!=-1) ans=min(ans,(n-(r-b[a[r]-m])));
        if(b[a[r]]==-1) b[a[r]]=r;
    }
    cout << ans<<endl;
}   
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
 
    cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

F. 3SUM

在这里插入图片描述
和的最后一位只与每一个数的最后一个数有关，打表找规律，分类讨论


#include <bits/stdc++.h>
 
#define pt2(x,y) cout <<(x)<<"--"<<(y)<<endl;
#define pt1(x) cout <<"#"<<(x)<<"#"<<endl;
#define hh cout << "\n"
#define rep(i,a,n) for(int i = (a); i <= int(n); i++) 
#define per(i,a,n) for(int i = (n); i >= int(a); i--) 
#define fi first;
#define se second;
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef double DB;
const int N=1e6+10,M=1e6+10,INF=0x3f3f3f3f;
 
int t=1;
 
void solve()
{
    int st[15]={0};
    int n;
    cin >> n;
    rep(i,1,n) 
    {
        int x;
        cin >> x;
        st[x%10]++;
    }
    if(st[1]>=3) 
    {
        cout <<"YES\n";
        return ;
    }
    rep(i,0,9)
    {
        if(st[i]>=2) 
        {
            rep(j,0,9) 
            if((2*i+j)%10==3)
            {
                if(st[j]==0 || i==j) continue;
                cout <<"YES\n";
                return;
            }
        }
    }
    rep(i,0,9) rep(j,i+1,9) rep(k,j+1,9) 
    {
        if(st[i]==0||st[j]==0||st[k]==0) continue;
        if((i+j+k)%10==3) 
        {
            cout <<"YES\n";
            return ;
        }
    }
    cout << "NO\n";
}   
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
 
    cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

G. 2^Sort

在这里插入图片描述
枚举左端点向右是否都满足条件

#include <bits/stdc++.h>
 
#define pt2(x,y) cout <<(x)<<"--"<<(y)<<endl;
#define pt1(x) cout <<"#"<<(x)<<"#"<<endl;
#define hh cout << "\n"
#define rep(i,a,n) for(int i = (a); i <= int(n); i++) 
#define per(i,a,n) for(int i = (n); i >= int(a); i--) 
#define fi first;
#define se second;
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef double DB;
const int N=1e6+10,M=1e6+10,INF=0x3f3f3f3f;
 
int t=1;
 
int a[N],b[N];
void solve()
{
    int n,k;
    cin >> n >> k;
    rep(i,1,n) cin >> a[i];
    int ans=0;
    rep(i,1,n-1)
    {
        if(a[i]<2*a[i+1]) b[i]=b[i-1]+1;
        else b[i]=0;
        if(b[i]>=k) ans++;  
    }
    cout << ans<<endl;
}   
int main()
{
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
 
    cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}