Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901).
Input
The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.OutputThe only line of the output will contain S modulo 9901.
Sample Input
2 3
Sample Output
15
Hint 2^3 = 8.
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).
题意:求AB的所有约数的和 % MOD (9901) 题意中有点问题,我们知道0是没有约数的,我觉得A、B应该都是>0的
思路:我们可以把A分解质因数(p1c1 * p2c2 * .... * pncn)B
约数和:(1 + p1 + p12 + ... + p1B*c1)* (1 + p2 + p22 + ... + p2B*c2)* .... * (1 + pn + pn2 + ... + pnB*cn) ( 排列组合问题)
这样我们可以看出这是多个等比数列乘积,可以用等比数列求和公式 (a1 *(1-qn))/(1-q),我们注意到这里有除法,但是同余模定理是对于加减乘的,那么我们可以利用费马小定理,
求出(1-q)的逆元,然后把除变成乘逆元
坑点:应为 9901 这个质数较小,很容易找到一个数x,(x-1)% MOD == 0 ,就说明这个数是没有逆元的(例217823),那么对于这种情况,我们不能用逆元算,你会发现这种情况下,
pn % MOD == 1 ((p-1)% MOD == 0)),这样(1 + pn + pn2 + ... + pnB*cn) == (1 % MOD + pn %MOD + pn2 %MOD + ... + pnB*cn %MOD) == B*cn+1
#include<iostream>
#include<cstdio>
#include<math.h>
using namespace std; const int maxn = 1e4;
const int mod = ;
int a,b;
int p[maxn];
int c[maxn];
int calc(int x)
{
int m= ;
int up = sqrt(x);
for(int i=;i<=up;i++)
{
if(x % i == )p[++m] = i,c[m] = ;
while(x % i == )x/= i,c[m]++;
}
if(x > )p[++m] = x,c[m] = ;
return m;
} typedef long long ll; ll qpow(ll a,ll b)
{
ll ans = ;
ll base = a;
while(b)
{
if(b&)ans = (ans * base)%mod;
base = (base * base)%mod;
b >>= ;
}
return ans;
}
int main()
{
scanf("%d%d",&a,&b);
int n = calc(a);
ll ans = ;
for(int i=;i<=n;i++)
{
if((-p[i])%mod == )
{
ans = (ans * (b * c[i]+ ))%mod;
continue;
}
ll Ni = qpow(-p[i],mod-);
ll tmp = -qpow(p[i],c[i]*b+);
ans = (ans * (tmp*Ni%mod+mod)%mod)%mod;
}
printf("%lld\n",ans);
}
还有一种写法,就是不用公式计算等比数列和,这样就避免了逆元的问题
sum(p,c) = (1 + p + p2 + ... + pk)
(1)c为奇数,sum(p,k)= sum(p,(k-1 )/2)*(1+p(k+1)/2)
sum(p,c) = (1 + p + p2 + ... + p(k-1)/2)+ (p(k+1)/2 + ... + pk) (c为奇数,加上0次幂,变成偶数,刚好可以分成两个等长的数列)
(2)c为偶数,sum(p,k)= sum(p,k/2-1)*(1+pk/2)+ pk
sum(p,c) = (1 + p + p2 + ... + pk/2-1)+ (pk/2 + ... + pk-1)+ pk (c+1是奇数)
#include<iostream>
#include<cstdio>
#include<math.h>
using namespace std; const int maxn = 1e4;
const int mod = ;
int a,b;
int p[maxn];
int c[maxn];
int calc(int x)
{
int m= ;
for(int i=;i*i<=x;i++)
{
if(x % i == )p[++m] = i,c[m] = ;
while(x % i == )x/= i,c[m]++;
}
if(x > )p[++m] = x,c[m] = ;
return m;
} typedef long long ll;
ll qpow(ll a,ll b)
{
ll ans = ;
ll base = a;
while(b)
{
if(b&)ans = (ans * base)%mod;
base = (base * base)%mod;
b >>= ;
}
return ans;
}
ll sum(ll p,ll c)
{
if(c == )return ;
if(c&)return ((+qpow(p,(c+)/))%mod*(sum(p,(c-)/)%mod))%mod;
else return ((+qpow(p,c/))%mod*(sum(p,c/-))%mod+qpow(p,c))%mod;
} int main()
{
scanf("%d%d",&a,&b);
int n = calc(a);
ll ans = ;
for(int i=;i<=n;i++)
{
ans = (ans * sum(p[i],c[i]*b))%mod;
}
printf("%lld\n",ans);
}