poj 1006 中国剩余定理解同余方程

其实画个图就明白了，

该问题就是求同余方程组的解：

n+d≡p (mod 23)

n+d≡e (mod 28)

n+d≡i (mod 33)

 #include "iostream"
 using namespace std;
 int a[],m[];
 int p,e,i,d,ans;

 int extend_gcd(int a,int b,int &x,int &y){
     if (b==){
         x=;y=;
         return a;
     }
     else{
         int r=extend_gcd(b,a%b,y,x);
         y=y-x*(a/b);
         return r;
     }
 }

 int CRT(int a[],int m[],int n)
 {
     int M=;
     for (int i=;i<=n;i++) M*=m[i];
     int ret=;
     for (int i=;i<=n;i++)
     {
         int x,y;
         int tm=M/m[i];
         extend_gcd(tm,m[i],x,y);
         ret=(ret+tm*x*a[i])%M;
     }
     return (ret+M)%M;
 }

 int main()
 {
     int T=;
     while (cin>>p>>e>>i>>d)
     {
         T++;
         if ((p==-)&&(e==-)&&(i==-)&&(d==-))
             break;
         a[]=p;     a[]=e;     a[]=i;
         m[]=;    m[]=;    m[]=;
         ans=CRT(a,m,);
         ans=ans-d;
         if (ans<)  ans+=;
         ans=ans%;
         if (ans==)     ans=;
         //Case 1: the next triple peak occurs in 1234 days.
         cout<<"Case "<<T<<": the next triple peak occurs in "<<ans<<" days."<<endl;
     }
     return ;
 }