Poj 1006 Chinese Residue Theorem

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

 

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

 

// Reference: http://blog.csdn.net/cyendra/article/details/38402869

For example, poj 1006

As a matter of fact, just draw a picture to understand,

This is the solution to the same equations:

N + D ≡ P (mod 23)

N + D ≡ E (mod 28)

N + d I (mod 33)

 

 1 #include "iostream" 2 using namespace std; 3 int a[5],m[5]; 4 int p,e,i,d,ans; 5  6 int extend_gcd(int a,int b,int &x,int &y){ 7     if (b==0){ 8         x=1;y=0; 9         return a;10     }11     else{12         int r=extend_gcd(b,a%b,y,x);13         y=y-x*(a/b);14         return r;15     }16 }17 18 int CRT(int a[],int m[],int n)19 {20     int M=1;21     for (int i=1;i<=n;i++) M*=m[i];22     int ret=0;23     for (int i=1;i<=n;i++)24     {25         int x,y;26         int tm=M/m[i];27         extend_gcd(tm,m[i],x,y);28         ret=(ret+tm*x*a[i])%M;29     }30     return (ret+M)%M;31 }32 33 int main()34 {35     int T=0;36     while (cin>>p>>e>>i>>d)37     {38         T++;39         if ((p==-1)&&(e==-1)&&(i==-1)&&(d==-1))40             break;41         a[1]=p;     a[2]=e;     a[3]=i;42         m[1]=23;    m[2]=28;    m[3]=33;43         ans=CRT(a,m,3);44         ans=ans-d;45         if (ans<0)  ans+=21252;46         ans=ans%21252;47         if (ans==0)     ans=21252;48         //Case 1: the next triple peak occurs in 1234 days.49         cout<<"Case "<<T<<": the next triple peak occurs in "<<ans<<" days."<<endl;50     }51     return 0;52 }

 

Poj 1006 Chinese Residue Theorem