Same remainder, Chinese Remainder Theorem

Cool: A between B (mod m), indicating a % m = B % m

The same formula:

Bytes --------------------------------------------------------------------------------------------------------------

 

The Chinese Remainder Theorem comes from a problem in Sun Tzu's computing Sutra:
I don't know the number of things, but the number of three is two, the number of five is three, and the number of seven is two. Ry?
In fact, this question is to solve such a homogeneous equations:
X limit 2 (mod 3)
X forward 3 (mod 5)
X limit 2 (mod 7) solving x

 

China Surplus theorem: solving the same-remainder Equations
N bought a [1] (mod m [1])
N bought a [2] (mod m [2])
......
N bought a [I] (mod m [I])
The equations have the following conditions:M [I]

As for the solution to this equation, Wikipedia explained in detail:

 1  int extend_gcd(int a,int b,int &x,int &y){ 2      if (b==0){ 3          x=1;y=0; 4          return a; 5      } 6      else{ 7          int r=extend_gcd(b,a%b,y,x); 8          y=y-x*(a/b); 9          return r;10      }11  }12  13  int CRT(int a[],int m[],int n)14  {15      int M=1;16      for (int i=1;i<=n;i++) M*=m[i];17      int ret=0;18      for (int i=1;i<=n;i++)19      {20          int x,y;21          int tm=M/m[i];22          extend_gcd(tm,m[i],x,y);23          ret=(ret+tm*x*a[i])%M;24      }25      return (ret+M)%M;26  }27 28  

 

// Reference: http://blog.csdn.net/cyendra/article/details/38402869
There is also an iterative solution to the Chinese Remainder Theorem:
Reference: http://scturtle.is-programmer.com/posts/19363.html

Same remainder, Chinese Remainder Theorem