「一本通 6.4 例 4」曹冲养猪(CRT)

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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() << "\n";
    return 0;
}

「一本通 6.4 例 4」曹冲养猪(CRT)

标签：ace   etc   pac   class   name   pair   include   air   efi   

原文地址：https://www.cnblogs.com/luoyibujue/p/10673305.html