也可以直接用等比树列求和公式，乘上逆元
分治：
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 using namespace std; 5 6 using ll = long long; 7 const int Mod = 9901; 8 int ksm(int a, int n){ 9 int res = 1; 10 while(n){ 11 if(n&1) res = (res%Mod * a % Mod) % Mod; 12 a = a % Mod * a % Mod; 13 n >>= 1; 14 } 15 return res; 16 } 17 18 int sum(int p, int n){ 19 if(n == 0) return 1; 20 if(n&1) return ((1 + ksm(p, (n+1)/2)) % Mod * sum(p, (n-1)/2)%Mod ) % Mod; 21 else return ((1 + ksm(p, n/2))% Mod * sum(p, n/2 - 1)%Mod * ksm(p, n) % Mod)% Mod; 22 } 23 24 int main(){ 25 int A, B; 26 int ans = 0; 27 cin >> A >> B; 28 for(int i = 2; i <= A; ++ i){ 29 int res = 0; 30 while(A % i == 0)res++, A /= i; 31 if(res) ans = (ans % Mod * sum(i, res * B) % Mod)%Mod; 32 } 33 cout << ans << endl; 34 return 0; 35 }
等比数列求和+逆元：
1 /* 2 Sn = (anq - a1)/(q-1) 3 由本题知q = i; 4 由于有n+1项； 5 Sn+1 = (an+1q - a1)/(q-1) 6 因为an = a1q^(n-1) 7 得 8 Sn+1 = (a1q^(n+1) - a1)/(q-1); 9 在由费马小定理 10 除以一个数并取模，等于乘上这个树得逆元，在取模; 11 求逆元用快速幂 12 ksm(x, Mod - 1) 13 */ 14 15 #include <iostream> 16 #include <algorithm> 17 #include <cstring> 18 using namespace std; 19 const int Mod = 9901; 20 using LL = long long; 21 LL ksm(LL a, LL n){ 22 LL res = 1; 23 while(n){ 24 if(n&1) res = (res%Mod * a%Mod)%Mod; 25 a = (a%Mod * a%Mod) % Mod; 26 n >>= 1; 27 } 28 return res; 29 } 30 31 int main(){ 32 LL A, B; 33 cin >> A >> B; 34 LL res = 1; 35 if(A == 0) res = 0; 36 for(int i = 2; i <= A; ++ i){ 37 if(A % i == 0){ 38 LL s = 0; 39 while(A % i == 0){ 40 s ++; 41 A /= i; 42 } 43 44 LL x = i % Mod; 45 if(i % Mod == 1)//避免出现为零的情况余数为1的时候逆元为零，不满足条件 46 res = res*(B*s + 1) % Mod; 47 else res = (res % Mod * (ksm(x, s * B + 1)-1) % Mod * ksm(x - 1, Mod - 2) %Mod) %Mod; 48 49 } 50 } 51 cout << res << endl; 52 return 0; 53 }