Sumdiv
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 15033 | Accepted: 3706 |
Description
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.
Output
The 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a^b的所有因子之和
解题思路：如果a是素数，我们可以知道a^b的所有因子是1，a，a^2，a^3，...，a^b，
则a^b的所有因子之和为1+a+a^2+a^3+a^4+...+a^b；
如果a不是素数，那么我们可以将a转换为由多个素数组成的合数即a=(p[0]^n[0])*(p[1]^n[1])*(p[2]^n[2])*...*(p[k]^n[k])，（p[0],p[1],p[2],...,p[n]都是素数），
那么a^b转换为p[0]^(n[0]*b) * p[1]^(n[1]*b) * p[2]^(n[2]*b) * ... * p[n]^(n[n]*b)，
则所有因子之和为[(1+p[0]+p[0]^2+...+p[0]^(n[0]*b) )*(1+p[1]+p[1]^2+...+p[1]^(n[1]*b) )*...*(1+p[n]+p[n]^2+...+p[n]^(n[n]*b) )
参考代码：
#include <iostream>
#include <stdio.h>
#include <string.h>
typedef long long ll;
using namespace std;
#define m 9901
ll mod_pow(ll a,ll b){ //快速幂
ll ans=1;
while (b>0){
if (b&1)
ans=ans*a%m;
a=a*a%m;
b>>=1;
}
return ans;
}
ll sum(ll p,ll n) //递归二分求 (1 + p + p^2 + p^3 +...+ p^n)%mod
{ //奇数二分式 (1 + p + p^2 +...+ p^(n/2)) * (1 + p^(n/2+1))
if(n==0) //偶数二分式 (1 + p + p^2 +...+ p^(n/2-1)) * (1+p^(n/2+1)) + p^(n/2)
return 1;
if(n%2) //n为奇数,
return (sum(p,n/2)*(1+mod_pow(p,n/2+1)))%m;
else //n为偶数
return (sum(p,n/2-1)*(1+mod_pow(p,n/2+1))+mod_pow(p,n/2))%m;
}
/*
ll sum(ll a,ll b){ //求 1+a^1+a^2+...+a^b的值
ll ans=mod_pow(a,b+1);
return (ans-1)/(a-1);
}
*/
int main(){
int a,b;
int p[10000],n[10000];
while (cin>>a>>b){
int k=0;
memset(n,0,sizeof(n));
for (int i=2;i<=a;i++){ //将a转换为(p[0]^n[0])*(p[1]^n[1])*(p[2]^n[2])*...*(p[k]^n[k])
if (a%i==0){
p[k]=i;
while (a%i==0){
n[k]++;
a/=i;
}
k++;
}
}
ll ans=1;
for (int i=0;i<k;i++){
ans=ans*sum(p[i],b*n[i])%m;
}
cout<<ans<<endl;
}
return 0;
}