題解

設答案為x，由題意得，同餘方程 => 。

然後求得最小的非負整數解 x 就是答案。

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;
void exgcd(LL a, LL b, LL& d, LL& x, LL& y)
{
    if (!b) { d=a; x=1; y=0; }
    else { exgcd(b, a%b, d, y, x); y -= x*(a/b); }
}
LL cal(LL a, LL b, LL c)
{
    LL GCD, x0, y0;
    exgcd(a, b, GCD, x0, y0);
    if (c%GCD) return -1;
    LL aa = a/GCD, bb = b/GCD;
    x0 *= c/GCD; y0 *= c/GCD;
    return (x0%bb+bb)%bb;
}
int main()
{
    LL A, B, C, k;
    while (~scanf("%lld%lld%lld%lld", &A, &B, &C, &k) && (A+B+C+k))
    {
        k = 1LL<<k;
        LL ans = cal(C, k, B-A);///A+C*x ≡ B (mod (2^k)) => C*x+k*y=B-A
        if (ans != -1) printf("%lld\n", ans);
        else printf("FOREVER\n");
    }
    return 0;
}
/*
1 3 2 4
1 5 2 4
1 2 4 3
0 0 0 0
*/