• Codeforces Round #643 (Div. 2)


    题目传送门

    还是视频题解

    我在这简单说一下D和F吧。
    关于D题的证明:
    我们首先如下构造:
    (1 1 1cdots s-(n-1))
    如果(n-1<s-(n-1)-1)显然存在一种构造方式使得另一个人gg,我们接下来就考虑(n-1geq s-(n-1)-1)(2n>s)的情况。

    • 现在有(s<2n),并且假设我们可以构造出一种合法的序列(a)和一个(k),使得序列找不到一段区间和为(k)或者(s-k)。我们现在考虑将(a)进行不断地拼接,那么上述条件就转化为不存在一段区间和为(k)
    • 我们截取(2k)段,因(a_i>0),所以我们容易发现有(2kn)个前缀,并且前缀和值各不相同,即有(2kn)个不同地值。
    • 显然和的最大值为(2ks),我们将(1...2ks)分组:([1,k+1],[2,k+2],...,[2ks-k,2ks]),由于限制条件,每组至多选择一个作为前缀和的值,所以至多(ks)个值。
    • 故有(2knleq ks)(2nleq s),与我们的前提不符,矛盾。

    所以就证明了当(s<2n)时无解的情况。
    感觉这证明方法还是挺巧妙的吧。。貌似很少见到这种类似的证明?

    (F)题也是一个很巧妙的题,需要将题目所给出的条件利用到最大化。

    • (frac{d}{2}leq ansleq 2d),我们可以发现至多可以不用知道两个次数为(1)的质因子或者一个次数至多为(2)的质因子。
    • 因此我们可以将范围减小到(700),我们假设求出了(X)中质因子不超过(700)的部分(X_1),若(X_1geq 3),那么(3cdot 700cdot 700cdot 700>10^9),显然此时满足上面可以不用管的情况,我们直接输出(2X_1);否则我们单独考虑一下,用上(d-7leq ansleq d+7)
    • 但是此时询问次数要超限,我们注意到询问(Qleq 10^{18}),所以对一个质因子(p_ileq 700)而言,我们可以两个质因子乘在一起询问,因为需要(9)个质数连乘才刚好到(10^9),所以我们这里需要(5)补。
    • 但是通过连乘找(700)的所有素数,貌似需要(18)次,所以我们这里还要再次进行缩小,大约缩到(600)多就行了。

    所以这个题也差不多做完了,就充分利用各种条件就行,神奇。


    代码如下:

    A. Sequence with Digits
    /*
     * Author:  heyuhhh
     * Created Time:  2020/5/16 19:38:21
     */
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <cmath>
    #include <set>
    #include <map>
    #include <queue>
    #include <iomanip>
    #include <assert.h>
    #define MP make_pair
    #define fi first
    #define se second
    #define pb push_back
    #define sz(x) (int)(x).size()
    #define all(x) (x).begin(), (x).end()
    #define INF 0x3f3f3f3f
    #define Local
    #ifdef Local
      #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
      void err() { std::cout << std::endl; }
      template<typename T, typename...Args>
      void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
      template <template<typename...> class T, typename t, typename... A> 
      void err(const T <t> &arg, const A&... args) {
      for (auto &v : arg) std::cout << v << ' '; err(args...); }
    #else
      #define dbg(...)
    #endif
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    //head
    const int N = 1e5 + 5;
     
    void run() {
        ll a, k;
        cin >> a >> k;
        --k;
        while (k--) {
            ll x = a;
            int Min = 10, Max = -1;
            while (x) {
                Min = min(Min, int(x % 10));
                Max = max(Max, int(x % 10));
                x /= 10;
            }
            if (Min == 0) break;
            a += Min * Max;
        }
        cout << a << '
    ';
    }
     
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
        cout << fixed << setprecision(20);
        int T; cin >> T; while(T--)
        run();
        return 0;
    }
    
    B. Young Explorers
    /*
     * Author:  heyuhhh
     * Created Time:  2020/5/16 19:44:50
     */
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <cmath>
    #include <set>
    #include <map>
    #include <queue>
    #include <iomanip>
    #include <assert.h>
    #define MP make_pair
    #define fi first
    #define se second
    #define pb push_back
    #define sz(x) (int)(x).size()
    #define all(x) (x).begin(), (x).end()
    #define INF 0x3f3f3f3f
    #define Local
    #ifdef Local
      #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
      void err() { std::cout << std::endl; }
      template<typename T, typename...Args>
      void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
      template <template<typename...> class T, typename t, typename... A> 
      void err(const T <t> &arg, const A&... args) {
      for (auto &v : arg) std::cout << v << ' '; err(args...); }
    #else
      #define dbg(...)
    #endif
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    //head
    const int N = 2e5 + 5;
     
    int n;
    int a[N];
     
    void run() {
        cin >> n;
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
        }
        sort(a + 1, a + n + 1);
        int ans = 0;
        int Min = -1, cnt = 0;
        for (int i = 1; i <= n; i++) {
            if (Min == -1) {
                cnt = 0;
            }
            Min = max(Min, a[i]);
            if (++cnt == Min) {
                Min = -1;
                ++ans;
            }
        }
        cout << ans << '
    ';
    }
     
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
        cout << fixed << setprecision(20);
        int T; cin >> T; while(T--)
        run();
        return 0;
    }
    
    C. Count Triangles
    /*
     * Author:  heyuhhh
     * Created Time:  2020/5/16 23:03:40
     */
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <cmath>
    #include <set>
    #include <map>
    #include <queue>
    #include <iomanip>
    #include <assert.h>
    #define MP make_pair
    #define fi first
    #define se second
    #define pb push_back
    #define sz(x) (int)(x).size()
    #define all(x) (x).begin(), (x).end()
    #define INF 0x3f3f3f3f
    #define Local
    #ifdef Local
      #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
      void err() { std::cout << std::endl; }
      template<typename T, typename...Args>
      void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
      template <template<typename...> class T, typename t, typename... A> 
      void err(const T <t> &arg, const A&... args) {
      for (auto &v : arg) std::cout << v << ' '; err(args...); }
    #else
      #define dbg(...)
    #endif
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    //head
    const int N = 1e6 + 5;
     
    int cnt[N];
     
    void run() {
        int A, B, C, D; cin >> A >> B >> C >> D;
        ++cnt[A + B], --cnt[B + B + 1];
        --cnt[A + C + 1], ++cnt[B + C + 2];
        for (int i = 1; i < N; i++) {
            cnt[i] += cnt[i - 1];
        }
        for (int i = 1; i < N; i++) {
            cnt[i] += cnt[i - 1];
        }
        ll ans = 0;
        for (int i = C + 1; i < N; i++) {
            int val = min(i - 1, D) - C + 1;
            ans += 1ll * val * cnt[i];
        }
        cout << ans << '
    ';
    }
     
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
        cout << fixed << setprecision(20);
        run();
        return 0;
    }
    
    D. Game With Array
    /*
     * Author:  heyuhhh
     * Created Time:  2020/5/16 20:24:54
     */
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <cmath>
    #include <set>
    #include <map>
    #include <queue>
    #include <iomanip>
    #include <assert.h>
    #define MP make_pair
    #define fi first
    #define se second
    #define pb push_back
    #define sz(x) (int)(x).size()
    #define all(x) (x).begin(), (x).end()
    #define INF 0x3f3f3f3f
    #define Local
    #ifdef Local
      #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
      void err() { std::cout << std::endl; }
      template<typename T, typename...Args>
      void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
      template <template<typename...> class T, typename t, typename... A> 
      void err(const T <t> &arg, const A&... args) {
      for (auto &v : arg) std::cout << v << ' '; err(args...); }
    #else
      #define dbg(...)
    #endif
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    //head
    const int N = 1e5 + 5;
     
    void run() {
        int n, s; cin >> n >> s;
        if (n == s) {
            cout << "NO" << '
    ';
            return;
        }
        int t = s - (n - 1);
        if (n - 1 >= t - 1 || s - n >= t) {
            cout << "NO" << '
    ';
            return;   
        }
        cout << "YES" << '
    ';
        for (int i = 1; i < n; i++) cout << 1 << ' ';
        cout << t << '
    ';
        cout << n << '
    ';
    }
     
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
        cout << fixed << setprecision(20);
        run();
        return 0;
    }
    
    E. Restorer Distance
    /*
     * Author:  heyuhhh
     * Created Time:  2020/5/16 22:21:11
     */
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <cmath>
    #include <set>
    #include <map>
    #include <queue>
    #include <iomanip>
    #include <assert.h>
    #define MP make_pair
    #define fi first
    #define se second
    #define pb push_back
    #define sz(x) (int)(x).size()
    #define all(x) (x).begin(), (x).end()
    #define INF 0x3f3f3f3f
    #define Local
    #ifdef Local
      #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
      void err() { std::cout << std::endl; }
      template<typename T, typename...Args>
      void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
      template <template<typename...> class T, typename t, typename... A> 
      void err(const T <t> &arg, const A&... args) {
      for (auto &v : arg) std::cout << v << ' '; err(args...); }
    #else
      #define dbg(...)
    #endif
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    //head
    const int N = 1e5 + 5;
     
    int n, A, R, M;
    int h[N];
     
    void run() {
        cin >> n >> A >> R >> M;
        M = min(M, A + R);
        for (int i = 1; i <= n; i++) {
            cin >> h[i];
        }
        auto f = [&] (int x) {
            ll cnt_a = 0, cnt_b = 0;
            for (int i = 1; i <= n; i++) {
                if (h[i] < x) cnt_a += x - h[i];
                if (h[i] > x) cnt_b += h[i] - x;   
            }
            ll cnt_c = min(cnt_a, cnt_b);
            cnt_a -= cnt_c;
            cnt_b -= cnt_c;
            return cnt_a * A + cnt_b * R + cnt_c * M;
        };
        int l = 0, r = INF, lmid, rmid;
        while (l < r) {
            lmid = l + ((r - l) / 3);   
            rmid = r - ((r - l) / 3);
            if (f(lmid) < f(rmid)) {
                r = rmid - 1;
            } else {
                l = lmid + 1;
            }
        }
        cout << f(l) << '
    ';
    }
     
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
        cout << fixed << setprecision(20);
        run();
        return 0;
    }
    
    F. Guess Divisors Count
    /*
     * Author:  heyuhhh
     * Created Time:  2020/5/17 11:13:54
     */
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <cmath>
    #include <set>
    #include <map>
    #include <queue>
    #include <iomanip>
    #include <assert.h>
    #define MP make_pair
    #define fi first
    #define se second
    #define pb push_back
    #define sz(x) (int)(x).size()
    #define all(x) (x).begin(), (x).end()
    #define INF 0x3f3f3f3f
    #define Local
    #ifdef Local
      #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
      void err() { std::cout << std::endl; }
      template<typename T, typename...Args>
      void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
      template <template<typename...> class T, typename t, typename... A> 
      void err(const T <t> &arg, const A&... args) {
      for (auto &v : arg) std::cout << v << ' '; err(args...); }
    #else
      #define dbg(...)
    #endif
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    //head
    const int N = 1e3 + 5;
    const int B = 630;
    const ll MAX = 1e18;
    int primes[N], tot;
    bool vis[N];
    void init() {
        for (int i = 2; i < N; i++) {
            if (!vis[i]) {
                primes[++tot] = i;
                vis[i] = true;
                for (ll j = 1ll * i * i; j < N; j += i) {
                    vis[j] = true;
                }
            }
        }
    }
     
    ll query(ll x) {
        cout << "? " << x << endl;
        ll t; cin >> t;
        return t;   
    }
    void answer(int x) {
        cout << "! " << x << endl;
    }
     
    void run() {
        int cnt = 0;
        vector <int> d;
        for (int i = 0, j = 1; i < 17; i++) {
            ll x = 1;
            while (x <= MAX / primes[j]) {
                x *= primes[j];
                ++j;
            }
            int t = query(x);
            for (int k = 1; k <= j; k++) {
                if (t % primes[k] == 0) {
                    d.push_back(primes[k]);
                }
            }       
        }
        
        auto gao = [&](int x) {
            int res = 1;
            while (1ll * res * x <= 1e9) {
                res *= x;
            }
            return res;
        };
        
    	auto calc = [&](int& t, int x) {
    		int cnt = 0;
    		while (t % x == 0) {
    			t /= x;
    			++cnt;
    		}
    		return cnt;
    	};
    	
        int x1 = 1, c = 1;
        for (int i = 0; i < sz(d); i += 2) {
            ll now = 1;
        	now *= gao(d[i]);
            int j = i + 1;
            if (j < sz(d)) now *= gao(d[j]);
            int t = query(now);
            x1 *= t;
            c *= (calc(t, d[i]) + 1);
            if (j < sz(d)) {
                c *= (calc(t, d[j]) + 1);   
            }
    	}
        if (x1 >= 4) answer(c << 1);
        if (x1 == 1) answer(8);
        if (x1 == 2 || x1 == 3) answer(9);
    }
     
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
        cout << fixed << setprecision(20);
        init();
        int T; cin >> T; while(T--)
        run();
        return 0;
    }
    
  • 相关阅读:
    Web开发(二)HTML
    Web开发(一)基础
    Python基础(十二)并发编程02
    Python基础(十一)并发编程01
    Python基础(十)socket编程
    Python基础(九)异常
    计算机基础
    Python基础(八)面向对象02
    jmeter使用正则表达式提取数据
    jmeter+ant 实现自动化接口测试环境配置
  • 原文地址:https://www.cnblogs.com/heyuhhh/p/12908798.html
Copyright © 2020-2023  润新知