poj1006 ( hdu1370 )：中国剩余定理裸题

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
#define MAX 200000000
#define ull unsigned long long
const int MAXN = 100011;
int a[3];
int m[3]={23,28,33};
int exgcd(int a,int b,int &x,int &y)
{
    if(b==0)
    {
        x=1;y=0;
        return a;
    }
    int r=exgcd(b,a%b,x,y);
    int t=x;
    x=y;
    y=(t-a/b*y);
    return r;
}
int china(int n)
{
    int M=1;
    int ans=0;
    int x,y,d;
    for(int i=0;i<n;i++)
    {
        M*=m[i];
    }
    for(int i=0;i<n;i++)
    {
        int mi=M/m[i];
        int x,y;
        d=exgcd(mi,m[i],x,y);
        ans=(ans+a[i]*mi*x)%M;
    }
    while(ans<0)
        ans+=M;
    return ans;
}
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         getchar();
         int p,e,d,n,x,f;
         int tt=0;
         int ans=21252;
         while(scanf("%d%d%d%d",&a[0],&a[1],&a[2],&d)&&(a[0]!=-1||a[1]!=-1||a[2]!=-1||d!=-1))
         {
             tt++;
             ans=china(3);
             while(ans<=d)
                 ans+=21252;
             printf("Case %d: the next triple peak occurs in %d days.\n",tt,ans-d);
         }
         while(t)
             printf("\n");
     }
    return 0;
}