POJ 2115 简单的模线性方程求解

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= =

 #include <cstdio>
 #include <cstring>

 using namespace std;
 #define ll long long
 ll ex_gcd(ll a , ll &x , ll b , ll &y)
 {
     if(b == ){
         x =  , y = ;
         return a;
     }
     ll ans = ex_gcd(b , x , a%b , y);
     int t = x;
     x = y , y = t - a/b*y;
     return ans;
 }
 int main()
 {
    // freopen("a.in" , "r" , stdin);
     int a , b , c , k;
     while(scanf("%d%d%d%d" , &a , &b , &c , &k)){
         if(a ==  && b ==  && c ==  && k == ) break;
         ll aa = c , bb =  , cc = b-a , x , y;
         for(int i= ; i<=k ; i++) bb = bb*;
         ll g = ex_gcd(aa , x , bb , y);
         if(cc % g != ){
             puts("FOREVER");
             continue;
         }
         ll kk = cc / g;
         x*=kk , y*=kk;
         aa/=g , bb/=g;
         if(x >= )
             x = x - x/bb*bb;
         else
             x = x - x/bb*bb+bb;
         printf("%I64d\n" , x);
     }
     return ;
 }