POJ 2115 C Looooops

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题目：http://poj.org/problem?id=2115

题意：对于C的for(i=A ; i!=B ;i +=C)循环语句，问在k位存储系统中循环几次才会结束。若在有限次内结束，则输出循环次数。否则输出死循环。

思路：这道题是一个扩展欧几里德算法的拓展，求单变元模线性方程即：Cx=(B-A)(mod 2^k)

扩展欧几里得算法和单变元模线性方程（传送门） + 比较详细的博客

代码：

/*ID: wuqi9395@126.comPROG:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define rep(i, n) for (int i = 0; i < n; i++)#define debug puts("===============")#define eps (1e-6)typedef long long ll;using namespace std;ll extend_gcd(ll a, ll b, ll &x, ll &y) {    if (b == 0) {        x = 1, y = 0;        return a;    }    else {        ll r = extend_gcd(b, a % b, y, x);        y -= x * (a / b);        return r;    }}vector<ll> line_mod_equation(ll a, ll b, ll n) {    ll x, y;    ll d = extend_gcd(a, n, x, y);    vector<ll> ans;    ans.clear();    if (b % d == 0) {        x = (x % n + n) % n;        x %= (n / d);        ans.push_back(x * (b / d) % (n / d));        //for (ll i = 1; i < d; i++) ans.push_back((ans[0] + i * n / d) % n);    }    return ans;}int main () {    //freopen("2.txt", "w", stdout);    ll a, b, c, k;    while(scanf("%lld%lld%lld%lld", &a, &b, &c, &k) , a || b || c || k) {        ll n = 1LL << k;        ll p = ((b - a) % n + n) % n;        vector<ll> ans = line_mod_equation(c, p, n);        if (ans.size() == 0) puts("FOREVER");        else printf("%lld\n", ans[0]);    }    return 0;}

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