CRT与扩展CRT(板子整理)

思路来源

https://www.cnblogs.com/zwfymqz/p/8425731.html

板子整理

1.CRT

中国剩余定理，利用逆元构造同余方程组的一个解，

要求方程组模数之间互质，x==c[i](mod m[i])

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll extgcd(ll a,ll b,ll &x,ll &y)
{
	ll d=a;
	if(b)d=extgcd(b,a%b,y,x),y-=(a/b)*x;
	else x=1,y=0;
	return d;
}
ll mul(ll u,ll v,ll p)
{
	return (u*v-(ll)((long double)u*v/p)*p+p)%p;
}
ll modpow(ll x,ll n,ll mod)
{
	ll res=1;
	for(;n;n/=2,x=mul(x,x,mod)%mod)
	if(n&1)res=mul(res,x,mod)%mod;
	return res;
}
ll CRT(ll r[],ll m[],int n)//x==r(mod m)方程组 共n项
{
	ll ans=0,M=1,x,y;
	for(int i=1;i<=n;++i)
	M*=m[i];
	for(int i=1;i<=n;++i)
	{
		ll mi=M/m[i];
		extgcd(mi,m[i],x,y);
		x=(x%m[i]+m[i])%m[i];
		ans=(ans+mul(mul(r[i],mi,M),x,M))%M;
	} 
	return (ans+M)%M;
}

2.扩展CRT

每次两两合并，扩展欧几里得构造合并后的方程的一个解

不需要模数之间互质，有解的充要条件与扩展欧几里得的有解充要条件相同

longlong+快速乘的板子交题迷之wa，只好换成__int128

#include<iostream>
#include<cstdio>
#include<map>
#define LL __int128
using namespace std;
typedef long long ll;
const int MAXN = 1e6+10;//最大方程数
LL p,q;
LL K, C[MAXN], M[MAXN], x, y;
LL gcd(LL a, LL b)
{
    return b == 0 ? a : gcd(b, a % b);
}
LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if (b == 0) {x = 1, y = 0; return a;}
    LL r = exgcd(b, a % b, x, y), tmp;
    tmp = x; x = y; y = tmp - (a / b) * y;
    return r;
}
LL inv(LL a, LL b)
{
    LL r = exgcd(a, b, x, y);
    while (x < 0) x += b;
    return x;
}
LL excrt(LL K)
{
    for (LL i = 1; i <= K; i++) 
	{
		scanf("%lld%lld",&p, &q);//x==c[i](mod m[i]) 
		M[i]=p; C[i]=q;
	}
    bool flag = 1;
    for (LL i = 2; i <= K; i++) 
	{
        LL M1 = M[i - 1], M2 = M[i], C2 = C[i], C1 = C[i - 1], T = gcd(M1, M2);
        if ((C2 - C1) % T != 0) {flag = 0; break;}
        M[i] = (M1 * M2) / T;
        C[i] = ( inv( M1 / T , M2 / T ) * (C2 - C1) / T ) % (M2 / T) * M1 + C1;
        C[i] = (C[i] % M[i] + M[i]) % M[i];
    }
    return flag ? C[K] : -1;
}
int main() 
{
    while (~scanf("%lld", &K)) 
	printf("%lld\n",(ll)(excrt(K))); 
    return 0;
}