解題關鍵：註意爆long long

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

其中，$m = \prod\limits_{j = 1}^k {{m_j}}$，$\forall 1 \le j \le k$，${M_j} = \frac{m}{{{m_j}}}$，$M_j^{ - 1}$是滿足${M_j}M_j^{ - 1} \equiv 1(\bmod m)$的一個整數

復雜度$O(n\log n)$

 1 #include<bits/stdc++.h>
 2 using
 
 namespace std;
 3 typedef long long ll;
 4 ll a[100],b[100];
 5 ll x,y;
 6 ll extgcd(ll a,ll b,ll &x,ll &y){
 7     int d=a;
 8     if(b){
 9         d=extgcd(b,a%b,y,x);
10         y-=a/b*x;
11     }else{
12         x=1,y=0;
13     }
14     return d;
15 }
16 int main(){
17     ll n,m=1
 
;
18     ll ans=0;
19     cin>>n;
20     for(int i=0;i<n;i++){ cin>>b[i]>>a[i];m*=b[i];}
21     for(int i=0;i<n;i++){
22         ll mi=m/b[i];
23         extgcd(mi,b[i],x,y);
24         x=(x+b[i])%b[i];
25         ans=(ans+mi*x*a[i]+m)%m;
26     }
27     cout<<ans<<endl;
 
28     return 0;
29 }

[51nod1079]中國剩余定理