CF1626F A Random Code Problem 题解

合集 - 题解(65)

1.[ARC126C] Maximize GCD 题解2023-08-142.[ARC126D] Pure Straight 题解2023-08-143.[ABC215D] Coprime 2 题解2023-08-144.CF793F Julia the snail 题解2023-08-145.CF1845D Rating System 题解2023-08-146.[JOI 2023 Final] Advertisement 2 题解2023-08-157.CF1513D GCD and MST 题解2023-08-158.「JOISC 2016 Day 2」雇佣计划 题解2023-08-159.[ABC134F] Permutation Oddness 题解2023-08-1510.CF1188D Make Equal 题解2023-08-1511.[ARC096E] Everything on It 题解2023-08-1512.[ARC117D] Miracle Tree 题解2023-08-1713.[JOISC 2014 Day3] 电压 题解2023-08-1714.CF803C Maximal GCD 题解2023-08-1715.CF1787E The Harmonization of XOR 题解2023-08-1716.CF1762D GCD Queries 题解2023-08-1717.[AGC061C] First Come First Serve 题解2023-08-1818.CF1575G GCD Festival 题解2023-08-1819.CF1656D K-good 题解2023-08-1920.CF1823F Random Walk 题解2023-08-2021.[ABC297G] Constrained Nim 2 题解2023-08-2022.CF1762E Tree Sum 题解2023-08-2123.CF1798E Multitest Generator 题解2023-08-2224.P7486 「Stoi2031」彩虹 题解2023-08-2225.CF1485C Floor and Mod 题解2023-08-2326.CF1681E Labyrinth Adventures 题解2023-08-2327.CF1023F Mobile Phone Network 题解2023-08-2428.CF258D Little Elephant and Broken Sorting 题解2023-08-2529.P7485 「Stoi2031」枫 题解2023-08-2530.[AGC030D] Inversion Sum 题解2023-08-2531.CF1815D XOR Counting 题解2023-08-2632.CF1864B Swap and Reverse 题解2023-08-3033.CF1864C Divisor Chain 题解2023-08-3034.CF1174E Ehab and the Expected GCD Problem 题解2023-08-3135.[AGC051B] Bowling 题解2023-09-01

36.CF1626F A Random Code Problem 题解2023-09-01

37.CF915G Coprime Arrays 题解2023-09-0238.[ABC318E] Sandwiches 题解2023-09-0339.[ABC318G] Typical Path Problem 题解2023-09-0340.CF838D Airplane Arrangements 题解2023-09-0341.[ABC319G] Counting Shortest Paths 题解2023-09-1142.[ABC313F] Flip Machines 题解2023-09-1543.[ABC320F] Fuel Round Trip 题解2023-09-1744.[ARC125B] Squares 题解2023-09-2745.[ARC124C] LCM of GCDs 题解2023-09-2746.[ARC135C] XOR to All 题解2023-09-2847.CF1874C Jellyfish and EVA 题解2023-10-0148.[ARC136C] Circular Addition 题解2023-10-0249.[ARC150D] Removing Gacha 题解2023-10-0350.高橋君 题解2023-10-0451.CF1842G Tenzing and Random Operations 题解2023-10-1052.[ARC167D] Good Permutation 题解2023-10-1653.[AGC061A] Long Shuffle 题解2023-10-2654.[ARC104B] DNA Sequence 题解2023-11-0255.[ARC104C] Fair Elevator 题解2023-11-0256.[ARC104D] Multiset Mean 题解2023-11-0257.[ARC104E] Random LIS 题解2023-11-0358.[ABC327G] Many Good Tuple Problems 题解2023-11-0559.[ARC105C] Camels and Bridge 题解2023-11-0660.[ARC105D] Let's Play Nim 题解2023-11-0661.[ARC105E] Keep Graph Disconnected 题解2023-11-0662.[ARC105F] Lights Out on Connected Graph 题解2023-11-0763.[ARC106E] Medals 题解2023-11-1364.[ARC107F] Sum of Abs 题解2023-11-1565.[ARC106F] Figures 题解2023-11-16

收起

题意

给定长度为 $n$ 的数组 $a$ 和一个整数 $k$ ，执行下面的代码：

long long ans = 0; //定义一个初始值为0的长整型变量
for(int i = 1; i <= k; i++) {
	int idx = rnd.next(0, n - 1); //生成一个介于0到n-1的随机数（含0和n-1）
  								 //每个数被选中的概率是相同的
	ans += a[idx];
	a[idx] -= (a[idx] % i);
}

求代码运行结束后 $ans$ 的期望值 $E$ 与 $n^k$ 相乘的结果，即 $E\times n^k$。

（$1\le n\le {10}^7,1\le k\le 17$）。

题解

考虑计算 $ans$ 的期望值 $E$，最后使之与 $n^k$ 相乘。设 $P=\mathrm{lcm}(1\sim k-1)$，可以看出数组 $a$ 中的元素 $a_i$ 最多会被减去 $a_{i}modP$。因此若将 $a_i$ 拆分为两部分：$a_i=\lfloor {\displaystyle \frac{a_i}{P}}\rfloor \times P+a_{i}modP$，前半部分的贡献是不变的，为

$$k\times {\displaystyle \frac{P\times \sum _{i=0}^{n-1}\lfloor {\displaystyle \frac{a_i}{P}}\rfloor }{n}}$$

下面着重考虑后半部分的贡献，因为值域较小，故考虑将其作为 $\mathtt{\text{DP}}$ 的状态。设 $f_{i,j}$ 为在第 $i$ 次操作后模 $P$ 为 $j$ 的期望元素个数。有转移

$$\begin{array}{rl}{f}_{i,j-(jmodi)}& \leftarrow {\displaystyle \frac{1}{n}}\times f_{i-1,j}\\ f_{i,j}& \leftarrow (1-{\displaystyle \frac{1}{n}})\times f_{i-1,j}\end{array}$$

计算完成 $f$ 数组后即可快速计算前半部分的贡献，为

$$\sum _{i=0}^{k-1}{\displaystyle \frac{\sum _{j}j\times f_{i,j}}{n}}$$

总复杂度 $\mathcal{O}(n+kP)$，可以通过本题。

Code

#include <bits/stdc++.h>

typedef long long valueType;
typedef std::vector<valueType> ValueVector;
typedef std::vector<ValueVector> ValueMatrix;

constexpr valueType MOD = 998244353;

template<typename T1, typename T2, typename T3 = valueType>
void Inc(T1 &a, T2 b, const T3 &mod = MOD) {
    a = a + b;

    if (a >= mod)
        a -= mod;
}

template<typename T1, typename T2, typename T3 = valueType>
T1 sub(T1 a, T2 b, const T3 &mod = MOD) {
    return a - b < 0 ? a - b + mod : a - b;
}

template<typename T1, typename T2, typename T3 = valueType>
T1 mul(T1 a, T2 b, const T3 &mod = MOD) {
    return (long long) a * b % mod;
}

template<typename T1, typename T2, typename T3 = valueType>
void Mul(T1 &a, T2 b, const T3 &mod = MOD) {
    a = (long long) a * b % mod;
}

template<typename T1, typename T2, typename T3 = valueType>
T1 pow(T1 a, T2 b, const T3 &mod = MOD) {
    T1 result = 1;

    while (b > 0) {
        if (b & 1)
            Mul(result, a, mod);

        Mul(a, a, mod);
        b = b >> 1;
    }

    return result;
}

valueType lcm(valueType a, valueType b) {
    return a / std::__gcd(a, b) * b;
}

int main() {
    valueType N, A0, X, Y, K, M;

    std::cin >> N >> A0 >> X >> Y >> K >> M;

    ValueVector source(N);

    source[0] = A0;

    for (valueType i = 1; i < N; ++i)
        source[i] = (source[i - 1] * X + Y) % M;

    valueType P = 1;

    for (valueType i = 2; i < K; ++i)
        P = lcm(P, i);

    ValueMatrix F(K, ValueVector(P, 0));

    for (auto const &iter: source)
        ++F[0][iter % P];

    valueType const InvN = pow(N, MOD - 2), reverseN = sub(1, InvN);

    for (valueType k = 1; k < K; ++k) {
        for (valueType i = 0; i < P; ++i) {
            Inc(F[k][i - i % k], mul(F[k - 1][i], InvN));
            Inc(F[k][i], mul(F[k - 1][i], reverseN));
        }
    }

    valueType ans = 0;

    for (valueType k = 0; k < K; ++k)
        for (valueType i = 0; i < P; ++i)
            Inc(ans, mul(F[k][i], i));

    for (auto const &iter: source)
        Inc(ans, mul(iter / P * P, K));

    Mul(ans, InvN);

    Mul(ans, pow(N, K));

    std::cout << ans << std::endl;

    return 0;
}