• Codeforces Round #590 (Div. 3)


    Codeforces Round #590 (Div. 3)

    A. Equalize Prices Again

    • 思路:水题

    • AC代码


    #include <algorithm>
    #include <iomanip>
    #include <iostream>
    #include <map>
    #include <math.h>
    #include <queue>
    #include <set>
    #include <sstream>
    #include <stack>
    #include <stdio.h>
    #include <string.h>
    #include <string>
    typedef long long ll;
    typedef unsigned long long ull;
    using namespace std;
    
    ll mult_mod(ll x, ll y, ll mod){
        return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
    }
    
    ll pow_mod(ll a, ll b, ll p){
        ll res = 1;
        while (b){
            if (b & 1)
                res = mult_mod(res, a, p);
            a = mult_mod(a, a, p);
            b >>= 1;
        }
        return res % p;
    }
    
    ll gcd(ll a, ll b){
        return b ? gcd(b, a % b) : a;
    }
    
    int q, n, a, sum, ans;
    
    int main(){
    #ifndef ONLINE_JUDGE
        freopen("my_in.txt", "r", stdin);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        cin >> q;
        while (q -- ){
            sum = 0;
            cin >> n;
            for (int i = 1; i <= n; i ++ ){
                cin >> a;
                sum += a;
            }
            if (sum % n == 0)
                ans = sum / n;
            else
                ans = sum / n + 1;
            cout << ans << "
    ";
        }
        return 0;
    }
    

    B. Social Network

    • 思路:模拟搞搞即可

    • AC代码


    #include <algorithm>
    #include <iomanip>
    #include <iostream>
    #include <map>
    #include <math.h>
    #include <queue>
    #include <set>
    #include <sstream>
    #include <stack>
    #include <stdio.h>
    #include <string.h>
    #include <string>
    typedef long long ll;
    typedef unsigned long long ull;
    using namespace std;
    
    ll mult_mod(ll x, ll y, ll mod){
        return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
    }
    
    ll pow_mod(ll a, ll b, ll p){
        ll res = 1;
        while (b){
            if (b & 1)
                res = mult_mod(res, a, p);
            a = mult_mod(a, a, p);
            b >>= 1;
        }
        return res % p;
    }
    
    ll gcd(ll a, ll b){
        return b ? gcd(b, a % b) : a;
    }
    
    int n, k, x;
    deque<int> q;
    set<int> st;
    
    int main(){
    #ifndef ONLINE_JUDGE
        freopen("my_in.txt", "r", stdin);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        cin >> n >> k;
        for (int i = 1; i <= n; i ++ ){
            cin >> x;
            if (st.count(x))
                continue;
            q.push_front(x);
            st.insert(x);
            if (q.size() == k + 1){
                st.erase(q.back());
                q.pop_back();
            }
        }
        cout << q.size() << "
    ";
        while (!q.empty()){
            cout << q.front() << " ";
            q.pop_front();
        }
        return 0;
    }
    

    C. Pipes

    • 思路:照着题意做

    • AC代码


    #include <algorithm>
    #include <iomanip>
    #include <iostream>
    #include <map>
    #include <math.h>
    #include <queue>
    #include <set>
    #include <sstream>
    #include <stack>
    #include <stdio.h>
    #include <string.h>
    #include <string>
    typedef long long ll;
    typedef unsigned long long ull;
    using namespace std;
    
    ll mult_mod(ll x, ll y, ll mod){
        return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
    }
    
    ll pow_mod(ll a, ll b, ll p){
        ll res = 1;
        while (b){
            if (b & 1)
                res = mult_mod(res, a, p);
            a = mult_mod(a, a, p);
            b >>= 1;
        }
        return res % p;
    }
    
    ll gcd(ll a, ll b){
        return b ? gcd(b, a % b) : a;
    }
    
    const int N = 2e5 + 10;
    
    int q, n;
    int mp[N][2];
    string s;
    
    bool dfs(int x, int y, int k, int n){
        if (x == n && y == 0)
            return false;
        else if (x == n && y == 1)
            return true;
        if (k <= 2 && mp[x][y] <= 2)
            return dfs(x + 1, y, 2, n);
        else if (k <= 2 && mp[x][y] > 2){
            if (y == 0)
                return dfs(x, 1 - y, 4, n);
            else
                return dfs(x, 1 - y, 5, n);
        }
        else if (k == 3){
            if (mp[x][y] > 2)
                return dfs(x, 1 - y, 4, n);
            else
                dfs(x + 1, y, 1, n);
        }
        else if (k == 4){
            if (mp[x][y] > 2)
                return dfs(x + 1, y, 6, n);
            else
                return false;
        }
        else if (k == 5){
            if (mp[x][y] > 2)
                return dfs(x + 1, y, 3, n);
            else
                return false;
        }
        else if (k == 6){
            if (mp[x][y] > 2)
                return dfs(x, 1 - y, 5, n);
            else
                return dfs(x + 1, y, 1, n);
        }
    }
    
    int main(){
    #ifndef ONLINE_JUDGE
        freopen("my_in.txt", "r", stdin);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        cin >> q;
        while (q -- ){
            cin >> n;
            cin >> s;
            for (int i = 0; i < n; i ++ )
                mp[i][0] = s[i] - '0';
            cin >> s;
            for (int i = 0; i < n; i ++ )
                mp[i][1] = s[i] - '0';
            if (dfs(0, 0, 1, n))
                cout << "YES
    ";
            else
                cout << "NO
    ";
        }
        return 0;
    }
    

    D. Distinct Characters Queries

    • 思路:用树状数组搞

    • AC代码


    #include <algorithm>
    #include <iomanip>
    #include <iostream>
    #include <map>
    #include <math.h>
    #include <queue>
    #include <set>
    #include <sstream>
    #include <stack>
    #include <stdio.h>
    #include <string.h>
    #include <string>
    typedef long long ll;
    typedef unsigned long long ull;
    using namespace std;
    
    ll mult_mod(ll x, ll y, ll mod){
        return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
    }
    
    ll pow_mod(ll a, ll b, ll p){
        ll res = 1;
        while (b){
            if (b & 1)
                res = mult_mod(res, a, p);
            a = mult_mod(a, a, p);
            b >>= 1;
        }
        return res % p;
    }
    
    ll gcd(ll a, ll b){
        return b ? gcd(b, a % b) : a;
    }
    
    const int N = 1e5 + 10;
    const int M = 26;
    
    int C[M][N];
    
    inline int lowbit(int x){
        return x & (-x);
    }
    
    inline void add(int pos, int x, int val){
        while (x < N){
            C[pos][x] += val;
            x += lowbit(x);
        }
    }
    
    inline int sum(int pos, int x){
        int ans = 0;
        while (x > 0){
            ans += C[pos][x];
            x -= lowbit(x);
        }
        return ans;
    }
    
    inline int query(int l, int r, int pos){
        return sum(pos, r) - sum(pos, l - 1);
    }
    
    int len, q, op, pos, l, r, ans;
    char c;
    string s;
    
    int main(){
    #ifndef ONLINE_JUDGE
        freopen("my_in.txt", "r", stdin);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        cin >> s;
        s.insert(0, " ");
        len = s.length() - 1;
        for (int i = 1; i <= len; i ++ )
            add(int(s[i] - 'a'), i, 1);
    // debug    cout << s << "
    ";
        cin >> q;
        while (q -- ){
            cin >> op;
            if (op == 1){
                cin >> pos >> c;
                add(int(s[pos] - 'a'), pos, - 1);
                s[pos] = c;
                add(int(s[pos] - 'a'), pos, 1);
            }
            else{
                ans = 0;
                cin >> l >> r;
                for (int i = 0; i < M; i ++ )
                    if (query(l, r, i) > 0)
                        ans ++ ;
                cout << ans << "
    ";
            }
        }
        return 0;
    }
    
    /*
    abacaba
    5
    2 1 4
    1 4 b
    1 5 b
    2 4 6
    2 1 7
    
    3
    1
    2
    */
    
    /*
    dfcbbcfeeedbaea
    15
    1 6 e
    1 4 b
    2 6 14
    1 7 b
    1 12 c
    2 6 8
    2 1 6
    1 7 c
    1 2 f
    1 10 a
    2 7 9
    1 10 a
    1 14 b
    1 1 f
    2 1 11
    
    5
    2
    5
    2
    6
    */
    
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  • 原文地址:https://www.cnblogs.com/Misuchii/p/11801514.html
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