简单的扩展欧几里得题
这里 2^k 不能自作聪明的用 1<<k来写 ， k >= 31时就爆int了 ， 即使定义为long long 也不能直接这样写
后来老老实实 for(int i=1 ; i<=k ; i++) bb = bb*2; 才过了= =
1 #include <cstdio> 2 #include <cstring> 3 4 using namespace std; 5 #define ll long long 6 ll ex_gcd(ll a , ll &x , ll b , ll &y) 7 { 8 if(b == 0){ 9 x = 1 , y = 0; 10 return a; 11 } 12 ll ans = ex_gcd(b , x , a%b , y); 13 int t = x; 14 x = y , y = t - a/b*y; 15 return ans; 16 } 17 int main() 18 { 19 // freopen("a.in" , "r" , stdin); 20 int a , b , c , k; 21 while(scanf("%d%d%d%d" , &a , &b , &c , &k)){ 22 if(a == 0 && b == 0 && c == 0 && k == 0) break; 23 ll aa = c , bb = 1 , cc = b-a , x , y; 24 for(int i=1 ; i<=k ; i++) bb = bb*2; 25 ll g = ex_gcd(aa , x , bb , y); 26 if(cc % g != 0){ 27 puts("FOREVER"); 28 continue; 29 } 30 ll kk = cc / g; 31 x*=kk , y*=kk; 32 aa/=g , bb/=g; 33 if(x >= 0) 34 x = x - x/bb*bb; 35 else 36 x = x - x/bb*bb+bb; 37 printf("%I64d " , x); 38 } 39 return 0; 40 }