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[5],m[5];
int p,e,i,d,ans;

int extend_gcd(int a,int b,int &x,int &y){
    if (b==0){
        x=1;y=0;
        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=1;
    for (int i=1;i<=n;i++) M*=m[i];
    int ret=0;
    for (int i=1;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=0;
    while (cin>>p>>e>>i>>d)
    {
        T++;
        if ((p==-1)&&(e==-1)&&(i==-1)&&(d==-1))
            break;
        a[1]=p;     a[2]=e;     a[3]=i;
        m[1]=23;    m[2]=28;    m[3]=33;
        ans=CRT(a,m,3);
        ans=ans-d;
        if (ans<0)  ans+=21252;
        ans=ans%21252;
        if (ans==0)     ans=21252;
        //Case 1: the next triple peak occurs in 1234 days.
        cout<<"Case "<<T<<": the next triple peak occurs in "<<ans<<" days."<<endl;
    }
    return 0;
}