topcoder srm 681 div1

problem1 link

二分答案。然后判断。将所有的机器按照$a_{i}$排序，$a_{i}$相同的按照$b_{i}$排序。用一个优先队列维护这些机器。这样对于第$i$个部分，拿出队列开始的机器来生产该部分；如果队列开头的机器生产的部分没用完，则将其左区间$a_{t}$设置为$a_{t}+1$然后重新塞到队列里。

problem2 link

首先由$A_{i}$得到$A_{i-1}$的方式为$A_{i-1}=((A_{i}+2^{50}-b)$^$a)$&$(2^{50}-1)$。然后就是一个数字一个数字暴力计算$B$值。总的累积复杂度是$O(n)$的。

problem3 link

令$f[i][j]$ 表示$vals[i]*vals[j]$对答案有贡献的概率。

令$dp[i][j][n]$表示$(i,j)$对答案有贡献且序列中总的元素个数为$n$的概率（那么$f[i][j]=dp[i][j][n]$）。考虑整个序列的长度从长度为 $n-1$转移到$n$。那么对于$dp[i][j][n]$来说，可以从下面三种情况转移而来：

（1）$dp[i-1][j-1][n-1]$: 在前面新增一个元素

（2）$dp[i][j-1][n-1]$: 在中间新增一个元素

（3）$dp[i][j][n-1]$: 在后面新增一个元素

code for problem1

#include <algorithm>
#include <queue>
#include <vector>

class FleetFunding {
 public:
  int maxShips(int m, const std::vector<int> &k, const std::vector<int> &a,
               const std::vector<int> &b) {
    int n = static_cast<int>(k.size());
    struct node {
      int L, R, cnt;

      node() = default;
      node(int L, int R, int cnt) : L(L), R(R), cnt(cnt) {}
      bool operator<(const node &A) const {
        if (L != A.L) return L > A.L;
        return R > A.R;
      }
    };
    auto Check = [&](int M) {
      if (M == 0) {
        return true;
      }
      std::priority_queue<node> Q;
      for (int i = 0; i < n; ++i) {
        Q.push(node(a[i], b[i], k[i]));
      }
      for (int i = 1; i <= m; ++i) {
        if (Q.empty()) {
          return false;
        }
        int sum = 0;
        while (sum < M) {
          if (Q.empty() || Q.top().L != i) {
            return false;
          }
          if (sum + Q.top().cnt < M) {
            sum += Q.top().cnt;
            Q.pop();
          } else {
            node p = Q.top();
            Q.pop();
            p.cnt -= M - sum;
            ++p.L;
            if (p.L <= p.R && p.cnt > 0) {
              Q.push(p);
            }
            break;
          }
        }
        while (!Q.empty() && Q.top().L == i) {
          node p = Q.top();
          Q.pop();
          ++p.L;
          if (p.L <= p.R) {
            Q.push(p);
          }
        }
      }
      return true;
    };

    int sum = 0;
    for (int i = 0; i < n; ++i) {
      sum += k[i];
    }
    int low = 0, high = sum / m;
    int result = 0;
    while (low <= high) {
      int M = (low + high) >> 1;
      if (Check(M)) {
        result = std::max(result, M);
        low = M + 1;
      } else {
        high = M - 1;
      }
    }
    return result;
  }
};

code for problem2

class LimitedMemorySeries2 {
 public:
  int getSum(int n, long long x0, long long a, long long b) {
    constexpr int kMod = 1000000007;
    constexpr long long M = (1ll << 50) - 1;

    auto NextX = [&](long long x) { return ((x ^ a) + b) & M; };

    auto PreX = [&](long long x) { return ((x + M + 1 - b) ^ a) & M; };

    auto Cal = [&](int id, long long x, int n) {
      int ll = id, rr = id;
      int result = 0;
      long long lx = x, rx = x;
      while (ll - 1 >= 0 && rr + 1 < n) {
        lx = PreX(lx);
        rx = NextX(rx);
        if (lx < x && x > rx) {
          ++result;
          --ll;
          ++rr;
        } else {
          break;
        }
      }
      return result;
    };

    int result = 0;
    long long t = x0;
    for (int i = 0; i < n; ++i) {
      result = (result + Cal(i, t, n)) % kMod;
      t = NextX(t);
    }
    return result;
  }
};

code for problem3

#include <vector>

class CoinFlips {
 public:
  double getExpectation(const std::vector<int> &vals, int prob) {
    const double p = prob / 1000000000.0;
    int n = static_cast<int>(vals.size());
    std::vector<double> p2(n + 1);
    p2[0] = 1.0;
    for (int i = 1; i <= n; ++i) {
      p2[i] = p2[i - 1] * (1 - p);
    }
    std::vector<std::vector<double>> prefix(n + 1, std::vector<double>(n + 1));
    std::vector<std::vector<double>> prefix_sum(n + 1,
                                                std::vector<double>(n + 1));
    for (int len = 3; len <= n; ++len) {
      for (int i = 1; i <= len; ++i) {
        prefix[len][i] = p2[i - 1] * p;
        if (i == 1) {
          prefix[len][i] += p2[len];
        }
        prefix_sum[len][i] = prefix_sum[len][i - 1] + prefix[len][i];
      }
    }
    std::vector<std::vector<std::vector<double>>> f(
        2, std::vector<std::vector<double>>(n + 1, std::vector<double>(n + 1)));
    int pre = 0, cur = 1;
    for (int len = 3; len <= n; ++len) {
      for (int i = 0; i <= n; ++i) {
        for (int j = 0; j <= n; ++j) {
          f[cur][i][j] = 0;
        }
      }
      for (int L = 1; L <= len; ++L) {
        for (int R = L + 2; R <= len; ++R) {
          f[cur][L][R] =
              prefix_sum[len][L - 1] * f[pre][L - 1][R - 1] +
              (prefix_sum[len][R - 1] - prefix_sum[len][L]) * f[pre][L][R - 1] +
              (1 - prefix_sum[len][R]) * f[pre][L][R];
          if (L + 2 == R) {
            f[cur][L][R] += p2[L] * p;
          }
        }
      }
      pre ^= 1;
      cur ^= 1;
    }
    double result = 0.0;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 2; j < n; ++j) {
        result += vals[i] * vals[j] * f[pre][i + 1][j + 1];
      }
    }
    return result;
  }
};