题意:求(A/B)%9973,可以转换为A*x(B关于9973的逆元,即Bx%9973=1)%9973,
由于9973是质数,所以可以用费马小定理求逆元
1 #include<iostream> 2 #include<string.h> 3 #include<string> 4 #include<sstream> 5 #include<vector> 6 #include<deque> 7 #include<map> 8 #include<algorithm> 9 #include<iomanip> 10 #include<math.h> 11 #include<set> 12 using namespace std; 13 14 typedef long long ll; 15 typedef unsigned long long ull; 16 const int mod = 9973; 17 18 ll quickmod(ll a,ll b) 19 { 20 ll ans=1; 21 while (b) 22 { 23 if (b & 1) 24 { 25 ans = ans * a%mod; 26 } 27 b = b / 2; 28 a = a * a%mod; 29 } 30 return ans; 31 } 32 int main() 33 { 34 int t; 35 cin >> t; 36 while (t--) 37 { 38 ll n, b; 39 cin >> n >> b; 40 cout<<n*quickmod(b,mod-2)%mod<<endl; 41 } 42 return 0; 43 }