45 分求调

#include<bits/stdc++.h>
#define int long long
using namespace std;
int t, p, e, r, d, x, y, ans, mul = 1, table[] = {0, 23, 28, 33}, a[4];
int exgcd(int a, int b, int &x, int &y)
{
	if (b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}
	int d = exgcd(b, a % b, x, y);
	int t = x;
	x = y;
	y = t - (a / b) * y;
	return d;
}
signed main()
{
	while (cin >> p >> e >> r >> d)
	{
		++ t;
		ans = 0;
		if (p == -1 && e == -1 && r == -1 && d == -1)
		{
			break;
		}
		a[1] = p;
		a[2] = e;
		a[3] = r;
		mul = 23 * 28 * 33;
		for (int i = 1; i <= 3; ++ i)
		{
			x = 0;
			y = 0;
			int k = mul / table[i];
			exgcd(k, table[i], x, y);
			ans += (x % mul * k % mul * a[i] % mul) % mul;
			ans %= mul;
		}
		while (ans <= d)
		{
			ans += mul;
		}
		cout << "Case " << t << ": the next triple peak occurs in " << (ans + mul) % mul - d << " days." << endl;
	}
	return 0;
}