POJ 1006 中国剩余定理

//Result:wizmann	1006	Accepted	720K	94MS	G++	1017B

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#include <cmath>



using namespace std;



#define input(x) cin>>x

#define print(x) cout<<x<<endl



const int cycle[]={23,28,33};

int mod[3];



int gcd(int a,int b)

{

	if(a<b) return gcd(b,a);

	else

	{

		if(a%b==0) return b;

		else return gcd(b,a%b);

	}

}



int lcm(int a,int b)

{

	return a*b/gcd(a,b);

}



void init()

{

	for(int i=0;i<3;i++)

	{

		int a=i;

		int b=(i+1)%3;

		int c=(i+2)%3;

		int l=lcm(cycle[a],cycle[b]);

		mod[c]=l;

		while(mod[c]%cycle[c]!=1) mod[c]+=l;

	}

}



int main()

{

	freopen("input.txt","r",stdin);

	int array[4];

	int cas=1;

	int MOD=cycle[0]*cycle[1]*cycle[2];

	init();

	while(input(array[0]) && array[0]>=0)

	{

		for(int i=1;i<4;i++) input(array[i]);

		long long ans=0;

		for(int i=0;i<3;i++) ans+=array[i]*mod[i];

		ans%=MOD;

		ans-=array[3];

		if(ans<=0) ans+=MOD;

		

		printf("Case %d: the next triple peak occurs in %lld days.\n",cas++,ans);

	}

	return 0;

}