P9816 少项式复合幂 题解

题目链接：少项式复合幂

注意到题目的模并不是很大，我们考虑两个核心的性质。

1. \(f(f(x)) \bmod p=f(f(x) \bmod p) \bmod p\)，证明直接代入 \(f(x)\) 进去得到：\(f(f(x))=a_0 \times f^0(x)+a_1\times f^1(x)...+a_n\times f^n(x) \bmod p=\)

\[\sum_{i=0}^{n}a_i \times f^{i}(x) \bmod p=\sum_{i=0}^{n}a_i \times (f^{i}(x) \bmod p) \bmod p，根据模运算的四则运算得证。
\]

1. \(f_{x_1+x_2}(x)=f_{x_2}(f_{x_1}(x))\)，这个很显然，进行 \(x_1\) 次函数迭代以后再进行 \(x_2\) 次函数迭代，总函数迭代次不变。

基于第二点，我们就可以倍增了，基于第一点，我们可以预处理出所有的 \(f(i) \bmod p\)，用于第二点的倍增。

倍增实际上就是将一个需要走 \(k\) 步的函数迭代，根据它的二进制分解为走 \(k_1+k_2+k_3...\) 步，而一个数的二进制数不会超过 \(\log{V_{max}}\) 级别。

参考代码

#include <bits/stdc++.h>

//#pragma GCC optimize("Ofast,unroll-loops")

#define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}

template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 1e5 + 10;
constexpr int PMax = 1e5;
int a[N], b[N];
int n, p, q;

inline int f(const ll x)
{
    ll ans = 0;
    forn(i, 1, n)
    {
        ll curr = qPow(x, b[i], p); //快速幂快速算x^b[i]
        ans = (ans + a[i] * curr) % p; //算出a*x^b[i]加上去
    }
    return ans;
}

constexpr int T = ceil(log2(1e7));
int go[N][T + 1]; //倍增预处理

inline void solve()
{
    cin >> n >> q >> p;
    forn(i, 1, n)cin >> a[i] >> b[i];
    forn(i, 0, PMax)go[i][0] = f(i);
    forn(j, 1, T)forn(i, 0, PMax)go[i][j] = go[go[i][j - 1]][j - 1];//倍增预处理
    while (q--)
    {
        int x, y;
        cin >> x >> y;
        x %= p;
        //二进制分解
        while (y)
        {
            int t = log2(y);
            x = go[x][t];
            y -= 1 << t;
        }
        cout << x << endl;
    }
}

signed int main()
{
    Spider
    //------------------------------------------------------
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
}

\[时间复杂度为 \ O(m\times p\log{b_{max}}+p\log{p}+q\log{y_{max}})
\]