poj 2115（模板题，扩展欧几里得算法求同余问题）

#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long
#include<cmath>
void gcd(ll a,ll b,ll&d,ll&x,ll&y){
    if(!b){
        d = a;x = 1;y = 0;
    }
    else{
        gcd(b,a%b,d,y,x);
        y -= x*(a/b);
    }
}
int main(){
    ll a,b,c,k;
    ll d,p,q,ans,x,m;
    
    while(scanf("%lld%lld%lld%lld",&a,&b,&c,&k)!=EOF){
        if(a==0&&b==0&&c==0&&k==0)break;
        m = (ll)pow((double)2.0,(double)k);
        gcd(c,m,d,p,q);
        x = b-a;
        if(x%d!=0)printf("FOREVER\n");
        else{
            ans = p*(x/d);
            m = m/d;
            ans = (ans%m+m)%m;
            printf("%lld\n",ans);
        }
    }
    return 0;
}