acwing 97. 约数之和

传送门

以下定理参考博客

#include

using namespace std;
typedef long long ll;
const ll mod = 9901;

ll a, b;
ll ksm(ll x, ll y)
{
    ll ans = 1;
    x%=mod;
    while(y)
    {
        if(y&1) ans = (ans*x)%mod;
        x = (x*x)%mod;
        y >>= 1;
    }
    return ans%mod;
}

ll work(ll p, ll k)  ///计算(p^0+p^1+....+p^k)
{
    if(k == 0) return 1;
    if(k%2 == 1) return (1+ksm(p, k/2+1))*work(p, k/2)%mod;
    else return (1+p*(work(p, k-1)))%mod;
}
int main()
{
    scanf("%lld%lld", &a, &b);
    
    ll ans = 1;
    for(ll i = 2; i*i <= a; ++i)
    {
        while(a%i == 0)
        {
            ll c = 0;
            while(a%i == 0) c++, a/=i;
            ans = (ans*work(i, b*c))%mod;
        }
    }
    if(a > 1) ans = (ans*work(a, b))%mod;
    if(a == 0) ans = 0; ///a等于0的时候特判答案为0
    printf("%lld\n", ans);
    return 0;
}