题目链接：
POJ 2115 C Looooops
题意：
有一个循环语句，变量从A开始，每次都要+C，终止条件是变量==B，每次累加都要mod 2^k.问循环最多执行几次？或者永远循环，
分析：
令K = 2^k.设循环执行了x次则：(A + C * x) mod K = B,也就是A + C * x = B + K * y.
移项可得：C * x - K * y = B - A.那么就是解这个不定方程的最小正整数解x。扩展欧几里德呀~
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;int k;
ll A, B, C, K;ll ex_gcd(ll a, ll b, ll& x, ll& y)
{if(b == 0) {x = 1, y = 0;return a;}ll d = ex_gcd(b, a % b, y, x);y -= a / b * x;return d;
}int main()
{while(~scanf("%lld%lld%lld%d", &A, &B, &C, &k) && (A || B || C || k)){if(C == 0){if(A == B) printf("0\n");else printf("FOREVER\n");continue;}K = (ll)1 << k;ll x, y, d;d = ex_gcd(C, K, x, y);if((B - A) % d != 0) printf("FOREVER\n");else {x = x * (B - A) / d;ll r = K / d;x = (x % r + r) % r;printf("%lld\n", x);}}return 0;
}