模板：中国剩余定理

Posted 2020-10-10

 1 //x=a[i](mod m[i]) 
 2 LL CRT(LL n,LL *a,LL *m){
 3     LL M=1,ans=0;
 4     for(int i=0;i<n;i++) M*=m[i];
 5     for(int i=0;i<n;i++){
 6         LL w=M/m[i];
 7         ans=(ans+w*inv(w,m[i])*a[i])%M;
 8     } 
 9     return (ans+M)%M;
10 }