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= =

 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\n" , x);

38     }

39     return 0;

40 }