POJ2115 C Looooops 一元模线性方程

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/*    题目描述：给出程序段for(int i = A ; i!= B ; i+= C)    statement ，给出A，B，C，k，已知该程序段的运算中出现              的数都是mod k之后的，问该程序段将被执行多少次结束，不会结束的话输出FOREVER                    方法：由题意列出方程有A + Cx = B （mod k），变形Cx = B - A（mod k），要求该模线性方程的最小正整数解*/#pragma warning(disable:4786)#pragma comment(linker, "/STACK:102400000,102400000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>#include<queue>#include<map>#include<set>#include<vector>#include<cmath>#include<string>#include<sstream>#define LL long long#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define mem(a,x) memset(a,x,sizeof(a))#define lson l,m,x<<1#define rson m+1,r,x<<1|1using namespace std;const int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;const double PI = acos(-1.0);const double eps=1e-8;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 = y - a/b * x;    return d;}LL linearmod(LL a , LL b , LL c)   //求解ax=b (mod c)的最小正整数解{    LL x , y;    LL d = ex_gcd(a , c , x , y);    if(b%d)        return -1;    x = b / d * x ;    LL m = c / d;    return (x % m + m)% m;}int main(){    LL a , b , c ;    int k;    while(scanf("%lld%lld%lld%d",&a , &b , &c , &k)==4 ){        if(!a && !b && !c && !k)        break;        LL real_mod = 1LL << k;        LL ans = linearmod(c ,b - a , real_mod);        if(ans == -1)            printf("FOREVER\n");        else            printf("%lld\n" , ans );    }    return 0;}
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