思路是带入 $x=1,0$ 后求出 $k,b$。

#include <bits/stdc++.h>
#define int long long
using namespace std;
int Mod, P;
string s;
stack <int> num;
stack <char> t;
int cal() {
	int y = num.top(); num.pop();
	int x = num.top(); num.pop();
	char op = t.top(); t.pop();
	if (op == '+') return (x + y) % Mod;
	if (op == '-') return (x - y + Mod) % Mod;
	if (op == '*') return x * y % Mod;
}
int level(char x) {
	if (x == '+' || x == '-') return 1;
	if (x == '*') return 2;
	return -1;
}
int work(int x) {
	while (!num.empty()) num.pop();
	while (!t.empty()) t.pop();
	int n = s.size(), now = 0;
	for (int i = 0; i < n; i++) {
		if (s[i] >= '0' && s[i] <= '9') {
		    now = (now * 10 % Mod + (s[i] - '0')) % Mod;
		    if (i == n - 1 || (s[i + 1] < '0' || s[i + 1] > '9'))
		        num.push(now);
		    now = 0;
		}
		else if (s[i] == 'x') num.push(x);
		else if (s[i] == '(') t.push(s[i]);
		else if (s[i] == ')') {
			while (!t.empty() && t.top() != '(') num.push(cal());
			t.pop();
		}
		else {
			while (!t.empty() && level(t.top()) >= level(s[i])) num.push(cal());
			t.push(s[i]);
		}
	}
	while (!t.empty()) num.push(cal());
	return num.top();
}
signed main() {
	string S;
	cin >> S >> P >> Mod;
	s = S; int b = work(0);
	s = S; int k = (work(1) - b + Mod) % Mod;
	for (int i = 0; i <= Mod; i++)
		if ((k * i % Mod + b) % Mod == P)
			return cout << i, 0;
	return 0;
}