莫欺少年穷，修仙之旅在这开始—>算法基础课题解

推导过程： $$x\bmod a_1=m_1,x\bmod a_2=m_2$$ $$x=k_1a_1+m_1=k_2a_2+m_2$$ $$k_1a_1-k_2a_2=m_2-m_1$$ $$令\;d=exgcd(a_1,a_2,k_3,k_4)$$ $$k_3a_1-k_4a_2=d$$ $$k_1=k_3 \cdot \dfrac{m_2-m_1}{d}$$ $$令\;k_1=k_1+k \cdot \dfrac{a_2}{d},k_2=k_2+k \cdot \dfrac{a_1}{d}$$ $$\begin{aligned} 故\;x&=k \cdot \dfrac{a_1a_2}{d}+k_1a_1+m_1\\\\&=ka_0+m_0\end{aligned}$$

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(!b)
    {
        x=1,y=0;
        return a;
    }

    LL d=exgcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}

LL mod(LL a,LL b)
{
    return (a%b+b)%b;
}

int main()
{
    int n;
    cin>>n;

    LL a1,m1;
    cin>>a1>>m1;

    for(int i=0;i<n-1;i++)
    {
        LL a2,m2;
        cin>>a2>>m2;

        LL k1,k2;
        LL d=exgcd(a1,a2,k1,k2);

        if((m2-m1)%d)
        {
            cout<<-1<<endl;
            return 0;
        }

        k1=mod(k1*(m2-m1)/d,a2/d);
        m1=k1*a1+m1;
        a1=a1/d*a2;
    }

    cout<<m1<<endl;

    return 0;
}