题面就是说，解方程 a + c * x Ξ b (mod 2k)
然后变个型：a + c * x - 2k * y = b - a。用exgcd求解即可。
刚开始我以为b - a还要正负讨论，但其实不用，因为正负数取摸结果不一样。
然后我调了半天是因为当a = b = c = k = 0的时候才退出，然后我写的是有一个为０就退出……
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 //const int maxn = ; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 ll a, b, c, k; 38 39 void exgcd(ll a, ll b, ll &x, ll &y, ll &d) 40 { 41 if(!b) d = a, x = 1, y = 0; 42 else {exgcd(b, a % b, y, x, d); y -= a / b * x;} 43 } 44 45 int main() 46 { 47 while(scanf("%lld%lld%lld%lld", &a, &b, &c, &k)) 48 { 49 if(!a && !b && !c && !k) break; 50 k = (ll)1 << k; //别忘了１得改成long long ,要不然1 << 32会爆 51 ll x, y, d; 52 exgcd(c, k, x, y, d); 53 if((b - a) % d) {puts("FOREVER"); continue;} 54 x = x * (b - a) / d; 55 ll t = k / d; 56 x = (x % t + t) % t; 57 write(x); enter; 58 } 59 return 0; 60 }