題意：給你A、B、C、k (1 <= k <= 32)，執(zhí)行下面的語句。 (0 <= A, B, C < 2^k)
long long cnt = 0;
for (variable = A; variable != B; variable += C)
cnt++;```
求循環(huán)結束后的cnt為多少，如果是死循環(huán)，就輸出"FOREVER"。其中0 <= variable < 2^k如果variable>=2^k、variable就需要取variable mod 2^k的余數。
**Sample Input**
> 3 3 2 16
3 7 2 16
7 3 2 16
3 4 2 16
0 0 0 0
**Sample Output**
> 0
2
32766
FOREVER
思路：根據循環(huán)語句可以得到A + Cx = B mod 2^k; 即Cx = (B - A) mod 2^k;
可以聯想到逆元，x是C關于2^k的乘法逆元，所以可以轉化公式為：Cx + (2^k)y = B - A，而a*x + b*y = c 有解的充要條件是c % gcd(a , b) == 0，此題也就迎刃而解了。
include <iostream>
using namespace std;
typedef long long ll;
ll extgcd(ll a, ll b, ll &x, ll &y) {
int d = a;
if (b != 0) {
d = extgcd(b, a % b, y, x);
y -= a / b * x;
}
else {
x = 1;
y = 0;
}
return d;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
ll A, B, C, k;
while (cin >> A >> B >> C >> k && A + B + C + k) {
k = (ll)1 << k;//注意強制轉換
ll x, y;
ll a = C;
ll b = k;
ll c = B - A;
ll gcd = extgcd(a, b, x, y);
if (c % gcd != 0) cout << "FOREVER" << endl;
else {
x *= c / gcd;
ll ans = (x % (b / gcd) + (b / gcd)) % (b / gcd);
cout << ans << endl;
}
}
return 0;
}
