预推免机试复习

Posted 2022-11-26 中二病没有蛀牙

https://acm.ecnu.edu.cn/contest/20/

A不等式

B
https://acm.ecnu.edu.cn/problem/3303/

https://acm.ecnu.edu.cn/contest/31/

预推免

https://acm.ecnu.edu.cn/contest/31/problem/A/
A. 吉吉木的野望

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main()

    ll d,k;
    cin>>d>>k;
    double sum = 0;
    ll an,sn;
    int cnt = 0;
    for(ll i = min(k,ll(1e7)) ;i >= 1;i--)
        an = 1 + d * (i - 1);
        // cout<<an<<endl;
        if(sum == 0) 
            sum = an;
        else
            sum = an + 1.0/sum;
        cnt++;
    
    sum =1.0/sum;
    printf("%.16lf\\n",sum);
    return 0;

B. 皇后问题

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+5;

ll visx[maxn];
ll visy[maxn];

ll k1[3*maxn],k2[3*maxn];
int main()

    ll n,x,y;
    ll cnt = 0;
    cin>>n;
    while(n--)
        cin>>x>>y;
        if(visx[x] != 0)
            cnt += visx[x];
        if(visy[y] != 0) 
            cnt += visy[y];    
        visx[x]++;
        visy[y]++;
        if(k1[y - x + maxn] != 0)
            cnt += k1[y-x + maxn];
        if(k2[x + y + maxn] != 0)
            cnt += k2[x+y +maxn];
        k1[y - x + maxn]++;
        k2[x + y + maxn]++;

    
    cout<<cnt<<endl;
    return 0;

C

过90%代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+5;
set<ll> vis;


ll slove(ll s)
    ll ans = 1;
    ll cnt = 0;
    ll x = s;
    for( ll i = 2; i <= sqrt(s);i++)    
        while(x % i == 0)
            x /=  i;
            cnt++;
        
        ans = ans*(cnt + 1);
        cnt = 0;
     
    if(x != 1) ans *= 2; 
    return ans;




int main()

    ll q,n;
    cin>>q>>n;
    ll sum = 0;
    if(q == 1)
        sum = slove(n);
    
    else
        for(ll a = 1 ;a  < n; a++)
            vis.clear();
            ll cnt = 0;
            for (ll x = 1; x <= (n - 1)/ a;x++ )
                ll by = n - a*x;
                for( ll i = 1 ;i<= sqrt(by);i++)
                    if(by % i ==0)
                        if(vis.find(i) == vis.end())
                            cnt++;
                            // cout<<a << " "<<i<<endl;
                        
                        vis.insert(i);
                        if(vis.find(by/i) == vis.end())
                            cnt++;
                            // cout<<a << " "<<by/i<<endl;
                        
                        vis.insert(by/i);
                        // vis[by /i] = 1;
                    
                
            
            sum += cnt;
        
    
    cout<<sum<<endl;
    return 0;

D. 我认识你
超时代码1：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 4e4+5;
vector<int> g[maxn];
set<int> vis;

int main()

    int n,m,u,v,q,s,t;
    cin>>n>>m;
    while(m--)
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
       
    cin>>q;
    while(q--)
        cin>>s>>t;
        vis.clear();
        int cnt = 0;
        for(auto c:g[s])
            vis.insert(c);
        
        for(auto c:g[t])
            if(vis.find(c) != vis.end()) cnt++;
            vis.insert(c);
        
        cout<<cnt<<endl;
    
    return 0;

超时代码2：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 4e4+5;
map<int, set<int> > g;


int main()

    int n,m,u,v,q,s,t;
    cin>>n>>m;
    while(m--)
        cin>>u>>v;
        g[u].insert(v);
        g[v].insert(u);
       
    cin>>q;
    while(q--)
        cin>>s>>t;
        int cnt = 0;
        set<int>::iterator it1 = g[s].begin();
        set<int>::iterator it2 = g[t].begin();
        // cout<<*it1<<endl;
        while (it1 != g[s].end() && it2 != g[t].end()) 
        
            if( *it1 < *it2)
                it1++;
            else if( *it1 == *it2)
                it1++;
                it2++;
                cnt++;
            
            else
                it2++;
        
        cout<<cnt<<endl;
        // cnt = g[s].size() + g[t].size() - cnt;
        // cout<<cnt<<endl;
    
    return 0;

正解：
用bitset二进制压缩优化
bitset类型在定义时就需要指定所占的空间，例如bitset<233>bit;
bitset类型可以用string和整数初始化（整数转化成对应的二进制）

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 4e4+5;
bitset<maxn> g[maxn];

int main()

    int n,m,u,v,q,s,t;
    cin>>n>>m;
    while(m--)
        cin>>u>>v;
        g[u][v] = 1;
        g[v][u] = 1;
       
    cin>>q;
    while(q--)
        cin>>s>>t;
        int cnt = 0;
        bitset <maxn> tmp = (g[s] & g[t]);
        cnt = tmp.count();
        cout<<cnt<<endl;
    
    return 0;