poj 1006 中国剩余定理解同余方程

#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;
}