【模板】中国剩余定理

#include <cstdio>
typedef long long ll;
int a[20], m[20], t;
ll qmul(ll a, ll b, ll m) {
	ll ans = 0;
	while (b) {
		if (b & 1) ans = (ans + a) % m;
		a = (a << 1) % m;
		b >>= 1;
	}
	return ans;
} 
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (b == 0) {
		x = 1, y = 0;
		return a;
	}
	ll d = exgcd(b, a % b, y, x);
	y -= a / b * x;
	return d;
}
ll CRT() {
	ll ans = 0;
	ll M = 1;
	ll x, y;
	for (int i = 1; i <= t; i++) M *= m[i];
	for (int i = 1; i <= t; i++) {
		ll Mi = M / m[i];
		exgcd(Mi, m[i], x, y);
		x = (x % m[i] + m[i]) % m[i]; //变成最小的解 
		ans = (ans + qmul(a[i] * Mi % M, x , M)) % M;
	} 
	return (ans + M) % M; //防止为负数 
}
int main() {
	scanf("%d", &t);
	for (int i = 1; i <= t; i++) {
		scanf("%d", a + i);
	} 
	for (int i = 1; i <= t; i++) {
		scanf("%d", m + i);
	} 
	printf("%lld",CRT());
	return 0;
}

#include <cstdio>
typedef long long ll;
int a[20], m[20], n;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (b == 0) {
		x = 1, y = 0;
		return a;
	}
	ll d = exgcd(b, a % b, y, x);
	y -= a / b * x;
	return d;
}

ll CRT() {
	ll ans = 0; 
	ll M = 1;
	ll t1, t2;
	for (int i = 1; i <= n; i++) M *= m[i];
	for (int i = 1; i <= n; i++) {
		ll Mi = M / m[i];
		exgcd(Mi, m[i], t1, t2);
		ans = (ans + a[i] * Mi % M * t1) % M;	
	}
	//return (ans % M + M) % M; //保证结果为正整数 
	return (ans + M) % M; //保证结果为正整数 
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d%d", m + i, a + i);		
	}
	printf("%lld", CRT());
	return 0;
}