poj 2115 C Looooops （扩展欧几里得）

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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;}

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