Chinese Remainder Theorem $ CRT $

Consider a congruence equation
\ [\ begin {cases} x \ equiv a_1 \ (mod \ b_1) \\ x \ equiv a_2 \ (mod \ b_2) \\\ quad \ quad \ quad \ vdots \\ x \ equiv a_n \ (mod \ b_n) \ end {cases} \]

Wherein \ (b_1, b_2, \ dots , b_n \) pairwise quality.

Order \ (m = \ Prod \ limits_ {I =. 1} ^ n-B_i \) , \ (M_i = \ FRAC {m} {B_i} \) , \ (t_i \) is the congruence equation \ (xM_i \ equiv 1 (mod \ b_i) \) \ of a solution

Then the solution for the equations above that \ (\ SUM \ limits_. 1 = {I}} ^ {n-a_i M_i t_i \) .

prove:

Consider \ (\ \ SUM) in a \ (I \) , for the \ (k = i \) apparently \ (a_i M_i t_i \ equiv a_k \ (MOD \ b_k) \) , if the \ (k \ not I = \) , then \ (a_i M_i t_i \ equiv 0 \ (MOD \ b_k) \) (as \ (M_i \) other (B \) \ product).

Therefore, the above (\ Sigma \) \ in \ (n-\) entry will meet each equation, would not affect the other, so that \ (x = \ sum \ limits_ {i = 1} ^ {n} a_i M_i t_i \) is the set of legal solutions.

QED.

Further, \ (X \) is the (mod \ m \) \ Solutions of the sense, i.e., solution set may be expressed as \ (\ + {X km: K \ in \ the Z \} \) .

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Board questions

\(Code\)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mult(ll a,ll b,ll c){
    if (b>a) swap(a,b);
    ll ret=0; for (;b;b>>=1,a=a*2%c) if (b&1) ret=(ret+a)%c;
    return ret;
}
void exgcd(ll a,ll b,ll &x,ll &y){
    if (b==0){x=1,y=0;return;}
    else{
        exgcd(b,a%b,x,y);
        ll tmp=x; x=y; y=tmp-a/b*x;
    }
}
ll a[110],b[110]; int n;
ll CRT(){
    ll m=1; for (int i=1;i<=n;i++)m*=b[i];
    for (int i=1;i<=n;i++) a[i]=(a[i]%m+m)%m;
    ll ans=0;
    for (int i=1;i<=n;i++){
        ll Mi=m/b[i],t,tmp;
        exgcd(Mi,b[i],t,tmp);
        t=(t%m+m)%m;
        ans=(ans+mult(Mi,mult(t,a[i],m),m))%m;
    }
    return ans;
}
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%lld",&a[i]);
    for (int i=1;i<=n;i++) scanf("%lld",&b[i]);
    printf("%lld\n",CRT());
    return 0;
}

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To be filled: Extending Chinese remainder theorem \ (ExCRT \)