http://acm.hdu.edu.cn/showproblem.php?pid=1370

中国剩余定理

已知(n+d)%23=a;   (n+d)%28=b;   (n+c)%33=i 
       使33×28×a被23除余1，用33×28×6=5544； 
       使23×33×b被28除余1，用23×33×19=14421； 
       使23×28×c被33除余1，用23×28×2=1288。 
      因此有（5544×p+14421×e+1288×i）% lcm(23,28,33) =n+d

又23、28、33互质，即lcm(23,28,33)= 21252;
      所以有n=（5544×p+14421×e+1288×i-d）%21252

本题所求的是最小整数解，避免n为负，因此最后结果为n= [n+21252]% 21252
那么最终求解n的表达式就是：

n=(5544*p+14421*e+1288*i-d+21252)%21252;

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<bitset>

using namespace std;

int main()

{

	int a , b , c , d ;

	int temp = 0 ;

	while( cin >> a >> b >> c >> d )

	{

		if( a == - 1 && b == -1 && c == -1 && d == -1 )

			break;

		int n = ( 5544 * a + 14421 * b + 1288 * c ) % 21252 ;

		int ans = n > d ? n - d : 21252 + n - d ;

		cout << "Case " << ++temp << ": the next triple peak occurs in " << ans << " days." << endl;

	}

	return 0 ;

}