Poj 1006 bioryhthms

Application of Chinese Remainder Theorem

 

23,28, 33

X = P (mod 23)

X = E (mod 28)

X = I (mod 33) (the equal sign of the two lines replaces the equal sign of the three lines ...)

 

The last X-D modulo 21252 again. Note that X-D may be negative or 0.

 

Since 23 28 33 has been fixed, pre-processing can be done first to calculate Mi and Mi-1, M * M-1 = 1 (mod Ni)

Code:

# Include <iostream> <br/> using namespace STD; <br/> struct node <br/> {<br/> int D; <br/> int X; <br/> int y; <br/>}; <br/> int M [4], Mm [4], a [4]; // A [I] x = m [I] (mod n [I]) <br/> int N [4] = {23,28, 33,0 }; <br/> struct node extended_eculid (int A, int B) <br/>{< br/> struct node T1, T2; <br/> If (B = 0) <br/>{< br/> t1.d = A; <br/> t1.x = 1; <br/> t1.y = 0; <br/> return T1; <br/>}< br/> T1 = extended_eculid (B, A % B); <br/> t2.d = t1.d; <br/> t2.x = t1.y; <br/> t2.y = t1.x-(A/B) * t1.y; <br/> return T2; <br/>}< br/> int mles (int, int B, int N) <br/>{< br/> struct node T; <br/> T = extended_eculid (A, n); <br/> return T. x * (B/T. d); <br/>}< br/> int main () <br/> {<br/> int I, ANS, Mul, CA = 0; <br/> Mul = 23*28*33; <br/> while (scanf ("% d", & A [0], & A [1], & A [2], & A [3]) <br/> {<br/> if (a [0] =-1 & A [1] =-1 & A [2] =-1 & & A [3] =-1) <br/> break; <br/> for (I = 0; I <3; I ++) <br/>{< br/> A [I] % = N [I]; <br/> M [I] = Mul/n [I]; <br/>}< br/> ans = 0; <br/> for (I = 0; I <3; I ++) <br/>{< br/> int Te; <br/> Te = mles (M [I], 1, n [I]); <br/> Te = tE + N [I]; <br/> ans + = tE * A [I] * m [I]; </P> <p >}< br/> ANS-= A [3]; <br/> ans = (ANS + 21252) % 21252; <br/> // ans % = 21252; <br/> If (ANS = 0) <br/> ans = 21252; <br/> Ca ++; <br/> printf ("case % d: The next Triple peak occurs in % d days. /n ", CA, ANS); <br/>}< br/> return 0; <br/>}