HDU 3579 Hello Kiki 中国剩余定理(合并方程

题意：

给定方程

res % 14 = 5

res % 57 = 56

求res

中国剩余定理裸题

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<queue>
#include<vector>
using namespace std;
#define N 10005
#define ll __int64
ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
//求一组解(x,y)使得 ax+by = gcd(a,b), 且|x|+|y|最小(注意求出的 x,y 可能为0或负数)。
//下面代码中d = gcd(a,b)
//可以扩展成求等式 ax+by = c,但c必须是d的倍数才有解,即 (c%gcd(a,b))==0
void extend_gcd (ll a , ll b , ll& d, ll &x , ll &y) {
	if(!b){d = a; x = 1; y = 0;}
	else {extend_gcd(b, a%b, d, y, x); y-=x*(a/b);}
}
ll inv(ll a, ll n) { //计算%n下 a的逆。如果不存在逆return -1;
	ll d, x, y;
	extend_gcd(a, n, d, x, y);
	return d == 1 ? (x+n)%n : -1;
}
ll n[N],b[N],len,lcm;
ll work(){
	for(ll i = 2; i <= len; i++) {
		ll A = n[1], B = n[i], d, k1, k2, c = b[i]-b[1];
		extend_gcd(A,B,d,k1,k2);
		if(c%d)return -1;
		ll mod = n[i]/d;
		ll K = ((k1*(b[i]-b[1])/d)%mod+mod)%mod;
		b[1] = n[1]*K + b[1];
		n[1] = n[1]*n[i]/d;
	}
	if(b[1]==0)return lcm;
	return b[1];
}
int main(){
	ll i,T,Cas=1;cin>>T;
	while(T--){
		cin>>len;
		lcm = 1;
		for(i=1;i<=len;i++) {
			cin>>n[i];
			lcm = lcm / gcd(lcm,n[i]) * n[i];
		}
		for(i=1;i<=len;i++)cin>>b[i];
		cout<<"Case "<<Cas++<<": ";
		cout<<work()<<endl;
	}
	return 0;
}

HDU 3579 Hello Kiki 中国剩余定理(合并方程