Poj 1006 Problem Solving report

 

1003,1004, 1005 too much water, not recordedCode. 1006 is also very simple. It is just a problem of undefined equations, because the problem is not clearly described, or because my mind is dizzy because I am not sleeping at noon, after reading the question several times, I did not understand what the meaning of the data is. I understood it only after reading the Chinese translation several times.

# Include <iostream>

Using namespace STD;

Int main ()

{

Int P, E, I, D;

Int start, Day;

Int COUNT = 1;

While (true)

{

Cin> P> E> I> D;

If (P =-1 & E =-1 & I =-1 & D =-1)

Break;

P = P % 23, E = E % 28, I = I % 33;

If (d <p)

Start = 0;

Else

Start = (D-P)/23 + 1; // Add 1 here, after the given date (excluding the given date)

For (int ix = start; ++ IX)

{

Day = P + 23 * IX;

If (day <E | day <I)

Continue;

If (day-D> 21252)

{

Cout <"case" <count ++ <": The next Triple peak occurs in" <21252 <"days." <Endl;

Break;

}

Else if (day-e) % 28 = 0 & (day-I) % 33 = 0)

{

Cout <"case" <count ++ <": The next Triple peak occurs in" <day-D <"days." <Endl;

Break;

}

}

}

Return 0;

}

The conclusion of this question is that if the Chinese Remainder Theorem is applied, it will be calculated in O (1) time, which is very powerful. For details, see

Http://blog.csdn.net/ChinaCzy/archive/2010/04/29/5544250.aspx