题意:求a^b的所有约数的和对9901取余。
分析:我们转化为a^b的所有质因子的等比数列的成积,例如100^1,转化为(1+2+4) * (1 + 5 + 25)。由于a^b的质因子与a的质因子相同,只是每个的数量是a的质因子的b倍。具体做法是先求所有素数,求a的所有质因子,对于每个质因子求num[i]*b+1项的等比数列。并求乘积。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
usingnamespace std;
#define maxn 10000
#define w 9901
int a, b;
boolis[maxn];
int prm[maxn];
int fac[maxn], num[maxn];
int getprm(int n)
{
int i, j, k =0;
int s, e = (int) (sqrt(0.0+ n) +1);
memset(is, 1, sizeof(is));
prm[k++] =2;
is[0] =is[1] =0;
for (i =4; i < n; i +=2)
is[i] =0;
for (i =3; i < e; i +=2)
if (is[i])
{
prm[k++] = i;
for (s = i *2, j = i * i; j < n; j += s)
is[j] =0;
}
for (; i < n; i +=2)
if (is[i])
prm[k++] = i;
return k;
}
int power(int a, int n)
{
int ret =1;
int m = a % w;
while (n)
{
if (1& n)
ret *= m;
ret %= w;
n >>=1;
m *= m;
m %= w;
}
return ret;
}
int cal(int d, int num)
{
d %= w;
if (num ==0)
return1;
if (num ==1)
{
return (d +1) % w;
}
int mid;
if (num &1)
{
mid = power(d, num /2+1);
return (mid +1) * cal(d, num /2) % w;
}
int x = power(d, num /2);
mid = power(d, num /2+1);
return ((mid +1) * cal(d, num /2-1) + x) % w;
}
int main()
{
//freopen("t.txt", "r", stdin);
scanf("%d%d", &a, &b);
int n = getprm(int(sqrt(a)) +1);
int temp =0;
for (int i =0; i < n; i++)
if (a % prm[i] ==0)
{
fac[temp] = prm[i];
while (a % prm[i] ==0)
{
a /= prm[i];
num[temp]++;
}
temp++;
}
n = temp;
int ans =1;
for (int i =0; i < n; i++)
ans = ans * cal(fac[i], num[i] * b) % w;
if (a !=1)
ans = ans * cal(a, b) % w;
printf("%d\n", ans);
return0;
}
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
usingnamespace std;
#define maxn 10000
#define w 9901
int a, b;
boolis[maxn];
int prm[maxn];
int fac[maxn], num[maxn];
int getprm(int n)
{
int i, j, k =0;
int s, e = (int) (sqrt(0.0+ n) +1);
memset(is, 1, sizeof(is));
prm[k++] =2;
is[0] =is[1] =0;
for (i =4; i < n; i +=2)
is[i] =0;
for (i =3; i < e; i +=2)
if (is[i])
{
prm[k++] = i;
for (s = i *2, j = i * i; j < n; j += s)
is[j] =0;
}
for (; i < n; i +=2)
if (is[i])
prm[k++] = i;
return k;
}
int power(int a, int n)
{
int ret =1;
int m = a % w;
while (n)
{
if (1& n)
ret *= m;
ret %= w;
n >>=1;
m *= m;
m %= w;
}
return ret;
}
int cal(int d, int num)
{
d %= w;
if (num ==0)
return1;
if (num ==1)
{
return (d +1) % w;
}
int mid;
if (num &1)
{
mid = power(d, num /2+1);
return (mid +1) * cal(d, num /2) % w;
}
int x = power(d, num /2);
mid = power(d, num /2+1);
return ((mid +1) * cal(d, num /2-1) + x) % w;
}
int main()
{
//freopen("t.txt", "r", stdin);
scanf("%d%d", &a, &b);
int n = getprm(int(sqrt(a)) +1);
int temp =0;
for (int i =0; i < n; i++)
if (a % prm[i] ==0)
{
fac[temp] = prm[i];
while (a % prm[i] ==0)
{
a /= prm[i];
num[temp]++;
}
temp++;
}
n = temp;
int ans =1;
for (int i =0; i < n; i++)
ans = ans * cal(fac[i], num[i] * b) % w;
if (a !=1)
ans = ans * cal(a, b) % w;
printf("%d\n", ans);
return0;
}