POJ - 2115 C Looooops（拓展欧几里得）

题目：

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

题意倒是很简单，就是求满足c*x-2^k*y=b-a的所有（x,y）中，x的最小非负值。

代码：

#include<iostream>
#include<stdio.h>
using namespace std;

long long x, y;

long long gcd(long long  a, long long b)
{
	if (a == 0 || b == 0)
	{
		x = (b == 0);
		y = (a == 0);
		return a + b;
	}
	long long r;
	if (a < 0)
	{
		r = gcd(-a, b);
		x *= -1;
		return r;
	}
	if (b < 0)
	{
		r = gcd(a, -b);
		y *= -1;
		return r;
	}
	if (a >= b)r = gcd(a%b, b);
	else r = gcd(a, b%a);
	y -= a / b*x;
	x -= b / a*y;
	return r;
}

int main()
{
	long long a, b, c, k, con = 1;
	while (cin >> a >> b >> c >> k)
	{
		if (k == 0)break;
		b -= a;
		k = (con << k);
		if (b == 0)cout << 0;
		else
		{
			long long g = gcd(c, -k);
			if (b%g)cout << "FOREVER";
			else cout << (x*b / g%k + k) % (k / g);
		}
		cout << endl;
	}
	return 0;
}

核心函数是拓展欧几里得函数gcd，不会的请点击打开我的博客