题解
设答案为x,由题意得,同余方程 =>
。
然后求得最小的非负整数解 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
*/