Poj2115 -- C looooops (Extended gcd)

C looooops

Time limit:1000 ms		Memory limit:65536 K
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 particle 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 2 163 7 2 167 3 2 163 4 2 160 0 0 0

Sample output

0232766FOREVER

Source

The initial value of the question is a. If you add c at each time, you can modulo 2 ^ K for the obtained value. When the value is B, the question will be asked at least how many times, if you have never stopped forever

From the question to the equation, if it is certain to stop, then after the loop X times, it will be equal to B (A + x * C) % (2 ^ K) = B. Obviously, expand the GCD question and obtain the equation. A + C * X-B = y * (2 ^ K), that is

Solve C * X-(2 ^ K) * Y = B-a and ask if there is any integer solution.

For a * X1 + B * Y1 = C, use the extension GCD to solve a * X1 + B * Y1 = GAD (a, B) = gcd (B, A % B) = A * y2 + B * (x2-a/B * Y2) to get X1 = Y2, Y1 = (X2 + A/B * Y2), using this equation, it can be pushed to B = 0, that is x = 1, y = 0, and the maximum number of public approx. A is returned layer by layer, and a * X1 + B * Y1 = GAD (, b). If C % (gcd (a, B) = 0, then a * X1 + B * Y1 = C has an integer, so that D = gcd (A, B, b), D = B/d. the smallest positive integer excludes x = (X % d + d) % d.

 

#include <cstdio>#include <cstring>#include <algorithm>#define LL __int64using 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 )<<k ;        gcd(c,k,d,x,y);        if( (b-a)%d )            printf("FOREVER\n");        else        {            x = (b-a)/d*x ;            k = k/d ;            x = (x%k+k)%k ;            printf("%I64d\n", x);        }    }    return 0;}