[Commonly used algorithm Summary - Chinese remainder theorem]

Chinese remainder theorem, also known as the Chinese remainder theorem (What is the name ah) is a method congruence group (see congruence) to solve the ancient Chinese. Number Theory is an important theorem. First, let a little! example! question! Now!

Notes: Three number abc, the remainder were m1 m2 m3,% calculated for the remainder of this year, && is the "and" operator. 

1 , respectively, to find two numbers can be divisible, and meet the minimum number is divisible by more than a third. 

K1 % B% C K1 == == 0 &&% A == K1 . 1 ; 

K2 % A% K2 == C == 0 &&% B == K2 . 1 ; 

K3 % A% B K3 == == 0 && % C == K3 . 1 ; 

2 , the three unknowns corresponding to the digital multiplier I plus up, subtract the number of three i.e. an integral multiple of the least common multiple of the results obtained. 

Answer = K1 × M1 + K2 × M2 + K3 × M3 - P × (A × B × C); 

P to satisfy answer > largest integer 0; 

or answer = (K1 × M1 + K2 × M2 + K3 × M3) % (a × b × c) ;

Let's prove little wave

　

　M2 = M1 * set * M * ...... Mn 

　　Mi = M / mi The 

　　set of inverse element Mi Mi ^ (- . 1 ) (mi The MOD) 

　　has Mi * Mi ^ (- . 1 ) ≡ . 1 (mi The MOD ) 

　　AI * Mi * Mi ^ (- . 1 ) ≡ai (mi the MOD) 

　　for all j not equal to I 

　　AI * Mi * Mi ^ (- . 1 ) ≡ 0 (MJ MOD) 

　　so the answer is that all AI * Mi * Mi ^ (- . 1 ) (P MOD) value

P3868 [TJOI2009] Guess

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 long long x,y;
 4 long long a[15],b[15];
 5 long long n;
 6 void exgcd(long long A,long long B)
 7 {
 8     if(B==0)
 9     {
10         x=1;
11         y=0;
12         return;
13     }
14     exgcd(B,A%B);
15     long long z=x;
16     x=y;
17     y=z-(A/B)*y;
18 }
19 long long fast(long long a1,long long b1,long long mod)
20 {
21     long long ans=0;
22     a1%=mod;
23     b1%=mod;
24     while(b1)
25     {
26         if(b1&1)
27         {
28             ans=(ans+a1)%mod;
29         }
30         b1>>=1;
31         a1=(a1+a1)%mod;
32     }
33     return ans;
34 }
35 long long china()
36 {
37     long long ans=0;
38     long long M=1;
39     for(long long i=1;i<=n;i++)
40         M*=b[i];
41     for(long long i=1;i<=n;i++)
42     {
43         long long m=M/b[i];
44         exgcd(m,b[i]);
45         while(x<0)
46             x+=b[i]; 
47         x%=b[i];
48         ans=(ans+fast(x,fast(m,(a[i]+M)%M,M),M)+M)%M;
49     }
50     return ans;
51 }
52 int main()
53 {
54     cin>>n;
55     for(long long i=1;i<=n;i++)
56     {
57         scanf("%lld",&a[i]);
58     }
59     for(long long i=1;i<=n;i++)
60     {
61         scanf("%lld",&b[i]);
62     }
63     cout<<china();
64     return 0;
65 }