poj2115--C Looooops(擴展gcd)

poj2115--C Looooops(擴展gcd)

C Looooops Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 17740 Accepted: 4600

Description

A Compiler Mystery: We are given a C-language style for loop of type

for (variable = A; variable != B; variable += C)

  statement;

I.e., a loop which starts by setting variable to value A and while variable is not equal to B, repeats statement followed by increasing the variable by C. We want to know how many times does the statement get executed for particular values of A, B and C, assuming that all arithmetics is calculated in a k-bit unsigned integer type (with values 0 <= x < 2^k) modulo 2^k.

Input

The input consists of several instances. Each instance is described by a single line with four integers A, B, C, k separated by a single space. The integer k (1 <= k <= 32) is the number of bits of the control variable of the loop and A, B, C (0 <= A, B, C < 2^k) are the parameters of the loop.

The input is finished by a line containing four zeros.

Output

The output consists of several lines corresponding to the instances on the input. The i-th line contains either the number of executions of the statement in the i-th instance (a single integer number) or the word FOREVER if the loop does not terminate.

Sample Input

3 3 2 16
3 7 2 16
7 3 2 16
3 4 2 16
0 0 0 0

Sample Output

0
2
32766
FOREVER

Source

題目大意初始值是a，每一次加c ，可以對得到的值取模2^k，在值為b時結束，問最少會執行多少次，如過一直不會停FOREVER

由題中大意的到等式，假設一定會停，那麼在循環x次後會等於b 也就是 （a + x*c）% （2^k）= b 。很明顯的擴展gcd的題目，得到等式， a + c*x - b = y*(2^k) 也就是

求解 c*x - (2^k)*y = b-a ,問有沒有整數解。

對於a*x1 + b*y1 = c ，用擴展gcd，求解 a*x1 + b*y1 = gad(a,b) = gcd(b,a%b) = a*y2 +b* (x2-a/b*y2)得到 x1 = y2 , y1 = ( x2 +a/b*y2 ) , 利用這個等式，可以推到b = 0 時，那是x = 1 , y = 0 ,最大公約數為a，在逐層返回，最終得到 a*x1 + b*y1 = gad(a,b) 的解，如果c%( gcd(a,b) ) == 0 那麼a*x1 + b*y1 = c有整數解，令d = gcd(a,b)，d = b / d.最小的正整數解圍 x = ( x%d +ｄ )%d.

#include 
#include 
#include 
#define LL __int64
using namespace std;
void gcd(LL a,LL b,LL &d,LL &x,LL &y)
{
    if(b == 0)
    {
        d = a ;
        x = 1 ;
        y = 0 ;
    }
    else
    {
        gcd(b,a%b,d,y,x);
        y = -y ; x = -x ;
        y += a/b*x ;
    }
    return ;
}
int main()
{
    LL k , a , b , c , d , x , y ;
    while(scanf("%I64d %I64d %I64d %I64d", &a, &b, &c, &k)!=EOF)
    {
        if(a == 0 && b == 0 && c == 0 && k == 0)
            break;
        k = ( (LL)1 )<