/**//***************************************
比较经典的中国剩余定理~~
使33×28被23除余1，用33×28×6=5544
使23×33被28除余1，用23×33×19=14421
使23×28被33除余1，用23×28×2=1288
（5544×p+14421×e+1288×i）%（23×28×33）=n+d
n=（5544×p+14421×e+1288×i-d）%（23×28×33）
****************************************/
#include <iostream>
#include <cstdio>
using std :: cin;
using std :: cout;
using std :: endl;
int get__( int a, int b, int c )
{
int ans;
int cnt = 1;
while( ( a * b * cnt ) % c != 1 )
{
cnt++;
}
return a * b * cnt;
}
int main()
{
int _p = get__( 28, 33, 23 );
int _e = get__( 23, 33, 28 );
int _i = get__( 23, 28, 33 );
int p, e, i, d;
int cas = 1;
while( cin >> p >> e >> i >> d )
{
if( p == e && e == i && i == d && d == -1 ) break;
int ans = ( _p * p + _e * e + _i * i - d ) % ( 23 * 28 * 33 );
if( ans <= 0 )
printf("Case %d: the next triple peak occurs in %d days.\n", cas++, 21252 - ans - 2 * d);
else
printf("Case %d: the next triple peak occurs in %d days.\n", cas++, ans);
}
return 0;
}
posted on 2010-10-02 20:18
Vontroy 阅读(542)
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