[51nod1079]中国剩余定理

解题关键：注意爆long long

$x \equiv {M_1}M_1^{ - 1}{a_1} + ... + {M_k}M_k^{ - 1}{a_k}(\bmod m)$

其中，$m = \prod\limits_{j = 1}^k {{m_j}}$，$\forall 1 \le j \le k$，${M_j} = \frac{m}{{{m_j}}}$，$M_j^{ - 1}$是满足${M_j}M_j^{ - 1} \equiv 1(\bmod m)$的一个整数

复杂度$O(n\log n)$

 

 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[100],b[100];
ll x,y;
ll extgcd(ll a,ll b,ll &x,ll &y){
    int d=a;
    if(b){
        d=extgcd(b,a%b,y,x);
        y-=a/b*x;
    }else{
        x=1,y=0;
    }
    return d;
}
int main(){
    ll n,m=1;
    ll ans=0;
    cin>>n;
    for(int i=0;i<n;i++){ cin>>b[i]>>a[i];m*=b[i];}
    for(int i=0;i<n;i++){
        ll mi=m/b[i];
        extgcd(mi,b[i],x,y);
        x=(x+b[i])%b[i];
        ans=(ans+mi*x*a[i]+m)%m;
    }
    cout<<ans<<endl;
    return 0;
}