zoukankan      html  css  js  c++  java
  • DISCO Presents Discovery Channel Code Contest 2020 Qual Task E. Majority of Balls

    Not able to solve this problem during the contest (virtual participation).

    The first observation is that if we can identify $N-1$ balls of which half is blue and the other half is red, then with these $N - 1$ balls we can identify the color of the rest $N+1$ balls.

    It's not hard to note that if there are more blue balls than red balls among balls numbered from $l$ to $l + N - 1$ but less among balls numbered from $l + 1$ to $l + N$, we then know that

    1. The $l$-th ball and the $l+N$-th ball must be of different colors.
    2. The $l$-th ball must be blue and the $l+N$-th ball must be red.
    3. There are equal number of blue and red balls among balls numbered from $l + 1$ to $l+N - 1$.

    The problem is whether such $l$ even exists? The answer is ,fortunately, YES.

    Here comes the second interesting observation:
    Let's assume without loss of generality, there are more blue balls than red balls among the first $N$ balls, then there must be more red balls among the last $N$ balls. So such $l$ as described above must exist, and we can find one using binary search which costs at most $log_2 N + 1$ queries. When the binary search finishes, we get $l$ and the color of the $l$-th and $l+N$-th balls.

    When $l$ is found, for each ball numbered from $1$ to $l - 1$ or from $l + N + 1$ to $2N$, we can know its color with a query. Note that exactly half of these $N - 1$ known balls are blue, so we can use these balls to identify color of balls numbered from $l + 1$ to $l + N -1$ in a similar way.

    code
    
    int main() {
      int n;
      cin >> n;
      auto ask = [&](int l) {
        cout << '?';
        for (int i = 0; i < n; i++) {
          cout << ' ' << l + i;
        }
        cout << endl;
        string res;
        cin >> res;
        return res.front();
      };
      char ml = ask(1);
      int l = 1, r = n + 1;
    

    while (r - l > 1) {
    int mid = l + (r - l) / 2;
    if (ask(mid) == ml) {
    l = mid;
    } else {
    r = mid;
    }
    }
    vector ans(2 * n + 1);
    ans[l] = ml;
    ans[l + n] = ml == 'R' ? 'B' : 'R';

    auto ask2 = [&](int pos) {
    cout << '?';
    for (int i = 1; i < n; i++) {
    cout << ' ' << l + i;
    }
    cout << ' ' << pos << endl;
    string res;
    cin >> res;
    return res.front();
    };
    // [l + 1, l + n)
    rng (i, 1, 2 * n + 1) {
    if (i < l || i > l + n) {
    ans[i] = ask2(i);
    }
    }
    auto ask3 = [&](int pos) {
    cout << '?';
    rng (i, 1, 2 * n + 1) {
    if (i < l || i > l + n) {
    cout << ' ' << i;
    }
    }
    cout << ' ' << pos << endl;
    string res;
    cin >> res;
    return res.front();
    };
    rng (i, l + 1, l + n) {
    ans[i] = ask3(i);
    }

    cout << "! ";
    for (int i = 1; i <= 2 * n; i++) {
    cout << ans[i];
    }
    cout << ' ';
    return 0;
    }

  • 相关阅读:
    storm 学习教程
    Scala 面向接口编程
    Scala 继承
    IntelliJ IDEA 代码检查规范QAPlug
    Spark入门实战系列
    IntelliJ Idea 常用快捷键 列表(实战终极总结!!!!)
    使用DOM解析XML文档
    栈结构Stack
    队列Queue ,双端队列Deque
    集合转换为数组toArray(),数组转换为集合asList()
  • 原文地址:https://www.cnblogs.com/Patt/p/11924674.html
Copyright © 2011-2022 走看看