POJ-----C Looooops

C Looooops
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 15282
 
Accepted: 3893
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
CTU Open 2004
 
주소:
http://poj.org/problem?id=2115
 
유클리드 확장
 
변환 하 는 식 은 c * x + d * y = (b - a) mod (d) 입 니 다.  d = 1LL < k (제목 의 뜻);
코드:

 1 #include<iostream>

 2 #include<cstdio>

 3 #define LL long long

 4 #include<cstring>

 5 using namespace std;

 6   LL x,y,q;

 7 void  exgcd(LL a, LL b)

 8 {

 9     if(b==0)

10     {

11         x=1,y=0,q=a;

12     }

13     else

14     {

15         exgcd(b,a%b);

16         LL temp=x;

17         x=y,y=temp-a/b*y;

18     }

19 }

20 

21 int main()

22 {

23   LL a,b,c,k;

24   //freopen("test.in","r",stdin);

25   //freopen("test.out","w",stdout);

26   while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&k),a+b+c+k)

27   {

28       LL temp=(LL)1<<k;

29       exgcd(c,temp);

30       if((b-a)%q)

31           printf("FOREVER
");

32       else

33       {

34           LL t=temp/q;

35           printf("%I64d
",(x*((b-a)/q)%t+t)%t);

36       }

37   }

38   return 0;

39 }