复习一下
扩展中国剩余定理
- 首先考虑两个同余方程
[x equiv a_1; mod; m_1\
x equiv a_2; mod; m_2
]
- 化成另一个形式
[x = n_1 * m_1 + a_1\
x = n_2 * m_2 + a_2
]
- 联立可得
[n_1 * m_1 + a_1 = n_2 * m_2 + a_2\
n_1 * m_1 - n_2 * m_2 = a_2 - a_1
]
- 有解的前提是
[gcd(m_1, m_2) |(a_2 - a_1)
]
- 设
[d = gcd(m_1, m_2)\
c = a_2 - a_1
]
- 则
[n_1 frac{m_1}{d} - n_2 frac{m_2}{d} = frac{c}{d}\
n_1 frac{m_1}{d} equiv frac{c}{d} mod frac{m_2}{d}
]
- 移项
[n_1 equiv frac{c}{d} * inv(frac{m_1}{d}, frac{m_2}{d}) mod frac{m_2}{d}\
n_1 = frac{c}{d} * inv(frac{m_1}{d}, frac{m_2}{d}) + y_1 * frac{m_2}{d}
]
然后(n_1)代入最上面的狮子可以得到
[x = (frac{c}{d} * inv(frac{m_1}{d}, frac{m_2}{d}) + y_1 * frac{m_2}{d}) * m_1 + a_1\
x = m_1 * frac{c}{d} * inv(frac{m_1}{d}, frac{m_2}{d}) + y_1 * frac{m_2 m_1}{d} + a_1\
x equiv m_1 * frac{c}{d} * inv(frac{m_1}{d}, frac{m_2}{d}) + a_1 mod frac{m_2 m_1}{d}
]
- 然后就是个新方程了
- 当然也适用于互质情况
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 22
#define mmp make_pair
using namespace std;
int read()
{
int nm = 0, f = 1;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
return nm * f;
}
ll gcd(ll a, ll b)
{
return !b ? a : gcd(b, a % b);
}
ll exgcd(ll a, ll b, ll &x, ll &y)
{
if(!b)
{
x = 1, y = 0;
return a;
}
else
{
ll d = exgcd(b, a % b, x, y);
ll tmp = x;
x = y;
y = tmp - a / b * y;
return d;
}
}
ll inv(ll a, ll p)
{
ll x, y;
ll d = exgcd(a, p, x, y);
if(d != 1) return -1;
return (x % p + p) % p;
}
ll a[M], b[M], n;
ll excrt()
{
ll a1 = a[1], m1 = b[1], a2, m2;
for(int i = 2; i <= n; i++)
{
a2 = a[i], m2 = b[i];
ll c = a2 - a1, d = gcd(m1, m2);
if(c % d) return -1;
ll k = inv(m1 / d, m2 / d);
m2 = m1 / d * m2;
a1 = m1 * c / d % m2 * k + a1;
m1 = m2;
a1 = (a1 % m1 + m1) % m1;
}
return a1;
}
int main()
{
n = read();
for(int i = 1; i <= n; i++) b[i] = read(), a[i] = read();
cout << excrt() << "
";
return 0;
}