poj 2115 C Looooops （扩展欧几里得）

http://poj.org/problem?id=2115

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

statement;

题意：在k位的系统内执行这个for循环，问这个循环多少次结束，k为的系统能表示的最大的数为2^k,超出2^k就从0开始

思路：由题意可得出公式(A + C*x) % 2^k = B,由同余模公式得(A%2^k + (C*x%2^k)) % 2^k = B

A + (C*x%2^k)) = B,    C*x%2^k) = B-A,  C*x ≡ (B-A) (mod) 2^k,即转换成标准的同余方程，根据扩展欧几里得求C的最小逆元就可以了

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 1002000
#define INF 0xfffffff
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))

typedef long long LL;

void Ex_gcd (LL a, LL b, LL &gcd, LL &x, LL &y)
{
if (!b)
{
x = 1;
y = 0;
gcd = a;
}
else
{
Ex_gcd (b, a%b, gcd, y, x);
y -= a/b*x;
}
}

int main ()
{///C*x ≡ (B-A) (mod 2^k)
LL A, B, C, k;
while (scanf ("%I64d %I64d %I64d %I64d", &A, &B, &C, &k), A+B+C+k)
{
LL a = C, b = 1LL<<k;
LL mod = (B-A+b)%b, x, y, gcd;

Ex_gcd (a, b, gcd, x, y);

if (mod%gcd)
{
puts ("FOREVER");
continue;
}

x = x * (mod/gcd) % b;
x = (x%(b/gcd) + b/gcd) % (b/gcd);

printf ("%I64d\n", x);
}
return 0;
}