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

 #include <cstdio>
 #include <cstring>

 using namespace std;
 #define ll long long
 const int N = ;
 int val[N];

 ll ex_gcd(ll a , ll b , ll &x , ll &y)
 {
     if(b == ){
         x= , y=;
         return a;
     }
     ll ans = ex_gcd(b,a%b,x,y);
     ll t=x;
     x=y,y=t-a/b*y;
     return ans;
 }

 ll inv(ll a , ll b , ll mod)
 {
     ll x , y;
     ll d = ex_gcd(b,mod,x,y);
     return a*x%mod;
 }

 int main()
 {
     int n,m;
     while(scanf("%d%d" , &n , &m ) == )
     {
         ll sum = ;
         for(int i= ; i<n ; i++){
             scanf("%d" , val+i);
             sum = (sum*val[i])%m;
         }
         for(int i= ; i<n ; i++){
             ll ans = (inv(sum , (ll)val[i] , m)+m)%m;
             if(i==) printf("%lld" , ans);
             else printf(" %lld" , ans);
         }
         printf("\n");
     }
     return ;
 }