1.前言

擴充套件中國剩餘定理好繞啊，很多地方取模都有講究，所以寫篇筆記來方便自己複習。

2.問題

給出一個二元組序列 \(\{(b_i,m_i)\}\)，要求找出一個 \(Q\)，滿足要求：

3.思路

1. 找到一個都滿足的解，將餘數修改為這個解（記錄下一個解）
2. 將兩個同餘方程的模數修改為 \(lcm (m_i,m_j)\)（合併， \(lcm (m_i,m_j\) 即為新方程的週期）
3. 繼續向下合併

具體式子：
合併兩個同餘方程
\(Q \equiv b_i (mod m_i)\)
\(Q \equiv b_j (mod m_j)\)

4.實現

思路相當簡單，但是為了防止溢位，我們需要及時的取模，而實現的取模特別有講究。

1.求解 \(x,y\) 的過程

由於我們合併的最終式子只有 \(x\)，所以我們可以不考慮 \(y\) 的值

條件為 \(x * m_i + b_i = y * m_j + b_j\)，等價於 \(x * m_i \equiv b_j - b_i \pmod {m_j}\)

所以 \(b_j - b_i\)

令 \(p = delta = \frac {lcm (m_i,m_j)}{m_i}\)

則 \(x\) 可以對 \(p\) 取模

2.合併的過程

令 \(M = lcm (m_i, m_j)\)，而 \(Q \equiv b_i + x * m_i \pmod {M}\)

則 \(Q(b_i + x* m_i)\) 可以對 \(M\) 取模

5.參考程式碼

#include <cstdio>
#include <iostream>
using namespace std;
#define LL long long
#define ULL unsigned long long

template <typename T> void read (T &x) { x = 0; T f = 1;char tem = getchar ();while (tem < '0' || tem > '9') {if (tem == '-') f = -1;tem = getchar ();}while (tem >= '0' && tem <= '9') {x = (x << 1) + (x << 3) + tem - '0';tem = getchar ();}x *= f; return; }
template <typename T> void write (T x) { if (x < 0) {x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0'); }
template <typename T> void print (T x, char ch) { write (x); putchar (ch); }
template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }

LL gcd (LL x, LL y) {
	if (y == 0) return x;
	else return gcd (y, x % y);
}
LL lcm (LL x, LL y) {
	return x / gcd (x, y) * y;
}
void exgcd (LL a, LL b, LL &x, LL &y) {
	if (b == 0) {
		x = 1; y = 0;
		return ;
	}
	exgcd (b, a % b, y, x);
	y -= (a / b) * x; 
}
LL mul (LL x, LL y, LL Mod) {
	LL res = 0;
	while (y) {
		if (y & 1) res = (res + x) % Mod;
		x = (x + x) % Mod; y >>= 1;
	}
	return res;
}
LL solve (LL a, LL b, LL c) {
	LL _gcd = gcd (a, b);
	c = (c % b + b) % b;
	if (c % _gcd != 0) return -1;
	
	LL x, y;
	exgcd (a, b, x, y);
	LL p = b / _gcd;
	x = (x % p + p) % p;
	x = mul (x, c / _gcd, p);
	return x;
}

const int Maxn = 1e5;

int n;
struct Eq {
	LL Mod, b;
}a[Maxn + 5];

int main () {
	read (n);
	for (int i = 1; i <= n; i++) {
		read (a[i].Mod); read (a[i].b);
		a[i].b %= a[i].Mod;
	}
	
	LL Mod = a[1].Mod, b = a[1].b;
	for (int i = 2; i <= n; i++) {
		LL res = solve (Mod, a[i].Mod, a[i].b - b);
		if (res == -1) {
			printf ("No Answer!");
			return 0;
		}
		LL M = lcm (Mod, a[i].Mod);
		b = (b + mul (res, Mod, M)) % M;
		Mod = M; 
	}
	printf ("%lld", b);
	return 0;
}