COJ 1163 乘法逆元的求解

乘法逆元就是求一个 a/b = c(mod m)在已知a%m , b%m 的条件下 求c的解

 1 #include <cstdio>

 2 #include <cstring>

 3 

 4 using namespace std;

 5 #define ll long long

 6 const int N = 100005;

 7 int val[N];

 8 

 9 ll ex_gcd(ll a , ll b , ll &x , ll &y)

10 {

11     if(b == 0){

12         x=1 , y=0;

13         return a;

14     }

15     ll ans = ex_gcd(b,a%b,x,y);

16     ll t=x;

17     x=y,y=t-a/b*y;

18     return ans;

19 }

20 

21 ll inv(ll a , ll b , ll mod)

22 {

23     ll x , y;

24     ll d = ex_gcd(b,mod,x,y);

25     return a*x%mod;

26 }

27 

28 int main()

29 {

30     int n,m;

31     while(scanf("%d%d" , &n , &m ) == 2)

32     {

33         ll sum = 1;

34         for(int i=0 ; i<n ; i++){

35             scanf("%d" , val+i);

36             sum = (sum*val[i])%m;

37         }

38         for(int i=0 ; i<n ; i++){

39             ll ans = (inv(sum , (ll)val[i] , m)+m)%m;

40             if(i==0) printf("%lld" , ans);

41             else printf(" %lld" , ans);

42         }

43         printf("\n");

44     }

45     return 0;

46 }