POJ2891 Strange Way to Express Integers【扩展中国剩余定理】

题目大意

就是模板。。。没啥好说的

思路

因为模数不互质，所以直接中国剩余定理肯定是不对的
然后就考虑怎么合并两个同余方程
(ans = a_1 + x_1 * m_1 = a_2 + x_2 * m_2)
(x_1 * m_1 + x_2 * m_2 = a _ 2 - a _ 1)(因为正负号没影响嘛)
然后就可以exgcd解出来(x_1, x_2)， 最后就可以得到(x' = a_1 + x_1 * m_1, m' = lcm(m_1, m_2))
然后就不停合并就可以了

---

//Author: dream_maker
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;
//----------------------------------------------
//typename
typedef long long ll;
//convenient for
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
//inf of different typename
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
//fast read and write
template <typename T>
void Read(T &x) {
  bool w = 1;x = 0;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') w = 0, c = getchar();
  while (isdigit(c)) {
    x = (x<<1) + (x<<3) + c -'0';
    c = getchar();
  }
  if (!w) x = -x;
}
template <typename T>
void Write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x; 
  }
  if (x > 9) Write(x / 10);
  putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 1e5 + 5;
ll n, a[N], m[N];
ll gcd(ll a, ll b) {
  return b ? gcd(b, a % b) : a;
}
void exgcd(ll a, ll b, ll &x, ll &y) {
  if (!b) {x = 1, y = 0; return;}
  exgcd(b, a % b, y, x);
  y -= a / b * x;
}
ll exCRT() {
  ll M = m[1], A = a[1];
  fu(i, 2, n) {
    ll d = gcd(M, m[i]), x, y;
    if ((a[i] - A) % d) return -1;
    exgcd(M, m[i], x, y);
    x *= (a[i] - A) / d;
    x = (x % (m[i] / d) + (m[i] / d)) % (m[i] / d);
    A += M * x, M = M / d * m[i], A %= M; 
  }
  if (A < 0) A += M;
  return A;
}
int main() {
  while (scanf("%lld", &n) != EOF){
    fu(i, 1, n) Read(m[i]), Read(a[i]);
    Write(exCRT());
    putchar('
');
  }
  return 0;
}