約數之和。

​​約數之和​​

題意：

思路：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod = 9901;

int cnt;
int pri[1000010];
int vis[1000010];
int ans[1000010];
int num[1000010];
ll qpow(ll a,ll b)
{
  ll res = 1;
  while(b)
  {
    if(b%2)res = res*a%mod;
    b = b>>1;
    a = a*a%mod;
  }
  return res;
}
void oula()
{
  for(int i = 2; i <= 1000000; i++)
  {
    if(!vis[i])pri[++cnt] = i;
    for(int j = 1; j <= cnt && pri[j]*i <= 1000000; j++)
    {
      vis[i*pri[j]] = 1;
    }
  }
}
int main()
{
  __int128 p;
  oula();
  ll a,b;
  cin>>a>>b;
  if(a == 0)
  {
    cout<<0<<endl;
    return 0;
  }
  int c = 0,now = a;
  for(int i = 1; i <= cnt; i++)
  {
    now = a;
    if(pri[i] > now)break;
    if(now%pri[i] == 0)ans[++c] = pri[i];
    while(now%pri[i]==0)
    num[c]++,now/=pri[i];
  }
  ll res = 1;
  for(int i = 1; i <= c; i++)
  {
    if((qpow(ans[i],b*num[i]+1)+mod-1)%mod != 0)
    res *= (qpow(ans[i],b*num[i]+1)+mod-1)%mod*qpow(ans[i]-1,mod-2)%mod;
    else res*=(num[i]*b+1)%mod; 
    res%=mod;
  }
  cout<<(res)%mod<<endl;
}