Looooops

题目描述：

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.

输入：

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.

输出：

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.

样例输入：

Sample Input
3 3 2 16 3 7 2 16 7 3 2 16 3 4 2 16 0 0 0 0

样例输出：

Sample Output
0 2 32766 FOREVER

题目大意：

首先来列方程：A + C * x - （2^k）* y = d

然后就得出了：C * x + （-2^k）* y = d

也就是：a = C ，b = 2^k（这里的b应该变为正数，把负号移到y上面去，最后我们要求的是x的值，与y没有关系，真不行用一次换元法，把-y换掉，最后那个特解求一下相反数就行了）
然后用扩展欧几里得求ax+by+gcd(a,b)

code:

#include<stdio.h>
long long exgcd(long long a,long long b,long long &x,long long &y)
{
	long long d=a;
	if(b!=0)
	{
		d=exgcd(b,a%b,y,x);
		y-=(a/b)*x;
	}
	else
	{
		x=1;
		y=0;
	}
	return d;
}
int main()
{
	long long a,b,c,x,y,k;
	while(~scanf("%lld%lld%lld%lld",&a,&b,&c,&k))
	{
		if(a==0&&b==c)break;
		long long what=(long long)1<<k;
		long long d=exgcd(c,what,x,y);
		what/=d;
		if((b-a)%d!=0)
		{
			printf("FOREVER\n");
		}
		else
		{
			x=(b-a)/d*x;
			printf("%lld\n",(x%what+what)%what);
		} 
	}
	return 0;
}