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。
- \(f(f(x)) \bmod p=f(f(x) \bmod p) \bmod p\)，证明直接代入 \(f(x)\) 进去得到：\(f(f(x))=a_0 \times f^0(x)+a_1\times f^1(x)...+a_n\times f^n(x) \bmod p=\)
- \(f_{x_1+x_2}(x)=f_{x_2}(f_{x_1}(x))\)，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();
}