63中国剩余定理

Posted 2020-10-26 passion27

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[500],r[500],k;
ll gcd(ll a,ll b){
    while(b!=0){
        ll t=a;
        a=b;
        b=t%b;
    }
    return a;
}
ll lcm(ll a,ll b){
    return a*b/gcd(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);
    ll t=x;x=y;y=t-a/b*y;
    return r;
}
ll inv(ll a,ll p){//a模p意义下的逆
    ll d,x,y;d=exgcd(a,p,x,y);
    return (x+p)%p==0?p:(x+p)%p;
}
ll CRT(ll a[],ll m[],ll n){
    ll M=1;
    ll ans=0;
    for(int i=1;i<=n;i++)
        M*=m[i];    //M为Πmi
        //并设Mi=M/mi，ti=Mi^-1表示Mi模mi意义下的倒数
        //即Miti≡1%mi；
    for(int i=1;i<=n;i++)
    {
        ll x,y;
        x=inv(M/m[i],m[i]);
        ans=(ans+M/m[i]*x*m[i])%M;
    }
    ans=(ans+M)%M;
    return ans;
}

inline ll read(){
    char c=getchar();ll s=0,f=1;
    while(c<\'0\'||c>\'9\'){
        if(c==\'-\') 
            f=-1;
        c=getchar();
    }
    while(c>=\'0\'&&c<=\'9\'){
        s=s*10+c-\'0\';
        c=getchar();
    }
    return s*f;
}
inline void write(ll x){
    if(x<0) {
        putchar(\'-\');
        x=-x;
    }
    if(x>9)
        write(x/10);
    putchar(x%10+\'0\');
}
int main(){
    k=read();
    for(int i=1;i<=k;i++)
    {
        a[i]=read();
        r[i]=read();
    }
    for(int i=2;i<=k;i++)
    {
        //两两合并
        ll r1=r[i-1],r2=r[i],a1=a[i-1],a2=a[i];
        ll g=gcd(a1,a2);
        if((r2-r1)%g)
        {
            printf("-1");
            return 0;
        }       
        a[i]=a1*a2/g;
        r[i]=((inv(a1/g,a2/g)*((r2-r1)/g)%(a2/g)*a1+r1)%a[i]+a[i])%a[i];
    }

    write(r[k]);
    return 0;
}