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  • CF 937

    A

    #include <bits/stdc++.h>
    #define PI acos(-1.0)
    #define mem(a,b) memset((a),b,sizeof(a))
    #define TS printf("!!!
    ")
    #define pb push_back
    #define inf 1e9
    //std::ios::sync_with_stdio(false);
    using namespace std;
    //priority_queue<int,vector<int>,greater<int>> que; get min
    const double eps = 1.0e-8;
    typedef pair<int, int> pairint;
    typedef long long ll;
    typedef unsigned long long ull;
    //const int maxn = 3e5 + 10;
    const int  maxm = 300;
    const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
    //priority_queue<int, vector<int>, less<int>> que;
    //next_permutation
    ll mod = 3e7;
    int num[606];
    int main()
    {
            int n;
            cin >> n;
            int cur;
            for (int i = 1; i <= n; i++)
            {
                    cin >> cur;
                    num[cur] = 1;
            }
            int anser = 0;
            for (int i = 1; i <= 600; i++)
            {
                    if (num[i])
                    {
                            anser++;
                    }
            }
            cout << anser << endl;
            return 0;
    
    }
    View Code

    B

    #include <bits/stdc++.h>
    #define PI acos(-1.0)
    #define mem(a,b) memset((a),b,sizeof(a))
    #define TS printf("!!!
    ")
    #define pb push_back
    #define inf 1e9
    //std::ios::sync_with_stdio(false);
    using namespace std;
    //priority_queue<int,vector<int>,greater<int>> que; get min
    const double eps = 1.0e-8;
    typedef pair<int, int> pairint;
    typedef long long ll;
    typedef unsigned long long ull;
    //const int maxn = 3e5 + 10;
    const int  maxm = 300;
    const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
    //priority_queue<int, vector<int>, less<int>> que;
    //next_permutation
    ll mod = 3e7;
    ll p, y;
    bool check(ll now)
    {
            for (int i = 2; i * i <= now && i <= p; i++)
            {
                    if (now % i == 0)
                    {
                            return 0;
                    }
            }
            return 1;
    }
    int main()
    {
            cin >> p >> y;
            int flag = 0;
            for (int i = y; i >= p + 1; i--)
            {
                    if (check(i))
                    {
                            cout << i << endl;
                            flag = 1;
                            break;
                    }
            }
            if (!flag)
            {
                    cout << -1 << endl;
            }
            return 0;
    }
    View Code

    C

    #include <bits/stdc++.h>
    #define PI acos(-1.0)
    #define mem(a,b) memset((a),b,sizeof(a))
    #define TS printf("!!!
    ")
    #define pb push_back
    #define inf 1e9
    //std::ios::sync_with_stdio(false);
    using namespace std;
    //priority_queue<int,vector<int>,greater<int>> que; get min
    const double eps = 1.0e-10;
    const double EPS = 1.0e-4;
    typedef pair<int, int> pairint;
    typedef long long ll;
    typedef unsigned long long ull;
    const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
    //priority_queue<int, vector<int>, less<int>> que;
    //next_permutation
    int main()
    {
            double k, d, t;
            cin >> k >> d >> t;
            ll k1 = k;
            ll d1 = d;
            double ans = 0;
            if (k1 % d1 == 0)
            {
                    printf("%.1f
    ", t);
                    return 0;
            }
            else
            {
                    double bei = (long long)(k / d) + 1.0;
                    //printf("bei: %.1f
    ", bei);
                    double xunhuan = bei * d;
                    //printf("xunhuan: %.1f
    ", xunhuan);
                    double ximie = xunhuan - k;
                    //printf("ximie: %.1f
    ", ximie);
                    double once = k + ximie / 2.0;
                    //printf("once: %.1f
    ", once);
                    double bei2 = (long long)(t / once);
                    //printf("bei2: %.1f
    ", bei2);
                    double remain = t - once * bei2;
                    //printf("remain: %.1f
    ", remain);
                    ans += bei2 * xunhuan;
                    //printf("ans: %.1f
    ",ans);
                    if (remain <= k)
                    {
                            ans += remain;
                    }
                    else
                    {
                            ans += k + (remain - k) * 2.0;
                    }
                    printf("%.1f
    ", ans);
            }
    }
    View Code

    D

    anser[i][j][0]表示偶数步能到达 [1]表示奇数部能到达

    每个点最多访问两次

    #include <bits/stdc++.h>
    #define PI acos(-1.0)
    #define mem(a,b) memset((a),b,sizeof(a))
    #define TS printf("!!!
    ")
    #define pb push_back
    #define inf 1e9
    //std::ios::sync_with_stdio(false);
    using namespace std;
    //priority_queue<int,vector<int>,greater<int>> que; get min
    const double eps = 1.0e-10;
    const double EPS = 1.0e-4;
    typedef pair<int, int> pairint;
    typedef long long ll;
    typedef unsigned long long ull;
    //const int maxn = 3e5 + 10;
    const int  maxm = 300;
    const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
    //priority_queue<int, vector<int>, less<int>> que;
    //next_permutation
    ll mod = 3e7;
    const int maxn = 200005;
    int anser[200005][3];
    int visit[200005];
    int flag = 0;
    int road[200005];
    vector<int> pla[200005];
    void dfs(int now, int x)
    {
            visit[now] = 1;
            anser[now][x % 2] = 1;
            road[x] = now;
            int to;
            int len = pla[now].size();
            if (len == 0 && (x % 2 == 0))
            {
                    cout << "Win" << endl;
                    for (int i = 1; i <= x; i++)
                    {
                            cout << road[i] << " ";
                    }
                    exit(0);
            }
            else
            {
                    for (int i = 0; i < len; i++)
                    {
                            to = pla[now][i];
                            if (visit[to] == 1)
                            {
                                    flag = 1;
                                    if (anser[to][(x + 1) % 2] == 1)
                                    {
                                            continue;
                                    }
                                    dfs(to, x + 1);
                            }
                            else
                            {
                                    if (anser[to][(x + 1) % 2] == 1)
                                    {
                                            continue;
                                    }
                                    dfs(to, x + 1);
                            }
                    }
            }
            visit[now] = 2;
    }
    int main()
    {
            for (int i = 1; i <= 200000; i++)
            {
                    for (int j = 0; j <= 1; j++)
                    {
                            anser[i][j] = -1;
                    }
            }
            int n, m;
            int num;
            int to;
            cin >> n >> m;
            for (int i = 1; i <= n; i++)
            {
                    scanf("%d", &num);
                    for (int j = 1; j <= num; j++)
                    {
                            scanf("%d", &to);
                            pla[i].pb(to);
                    }
            }
            int aim;
            cin >> aim;
            dfs(aim, 1);
            if (flag)
            {
                    cout << "Draw" << endl;
                    return 0;
            }
            cout << "Lose" << endl;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/Aragaki/p/8607717.html
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