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;
}