/*
http://poj.org/problem?id=1845
Time Limit: 1000MS
Memory Limit: 30000K
Total Submissions: 11846
Accepted: 2835
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).
Source
Romania OI 2002
解析:求A^B得所有约数之和,即因子之和,并其结果模去9901
思路:
递归+分治
这里运用了整数划分定理和求约数和公式
1.整数唯一分解定理:任何正整数有且只有一种方式写出素因子的乘积表达式
即:A=(p1^k1)*(p2*k2)*(p3^k3)*.....(pn*kn);
(pi均为素数)
2.约分和公式:(从定理1以及排列组合原理可得理解)
s=(1+p1^1+....+p1^k1)*(1+p2^1+.....+p2^k2).....(1+pn^1+....+pn^kn);
3.利用递归分治法求s:令sum=1+p^1+p^2+...p^n则
(1)n%2==1:sum=(1+p+p^2+....p^(n/2))*(1+p^(n/2+1));
(2)n%2==0:sum=(1+p+p^2+...p^(n/2-1))*(1+p^(n/2+1))+p^(n/2)
Problem
Result
Memory
Time
Language
Length
Submit Time
Accepted
1836 KB
47 ms
C++
1631 B
2013-07-24 16:56:20
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include <iostream>
using namespace std;
const int maxn=100000+10;
bool isprime[maxn];
long long prime[maxn],k[maxn],p[maxn];//
long long cnt,t;
int n,m;
const int mod=9901;
void makeprime()//构造素数表
{
memset(prime,0,sizeof(prime));
memset(isprime,true,sizeof(isprime));
t=0;
for(int i=2;i<=maxn;i++)
{ if(isprime[i])
prime[t++]=i;
for(int j=0;j<t&&i*prime[j]<=maxn;j++)
{
isprime[i*prime[j]]=false;
if(i%prime[j]==0)
break;
}
}
// for(int i=0;i<t;i++)
// printf("%d\n",prime[i]);
}
void dividedata()//整数划分
{
memset(k,0,sizeof(k));
int i,j;
for(i=0;i<t&&prime[i]<=n;i++)
{
if(n%prime[i]==0)
{p[++cnt]=prime[i];
while(n%prime[i]==0)
{k[cnt]++;
n=n/prime[i];
}
}
}
if(n!=1)//还需要对余下的数进行判断
{
p[++cnt]=n;
k[cnt]=1;
}
}
int pow_mod(long long a,long long b)
{
if(b==0)
return 1;
int x=pow_mod(a,b/2);
long long ans=(long long )x*x%mod;
if(b%2==1)
ans=ans*a%mod;
return (int)ans;
}
int sum_mod(long long a,long long b)
{
if(b==0)
return 1;
long long s1;
long long s2;
if(b%2==1)
{
s1=sum_mod(a,b/2);
s2=pow_mod(a,b/2+1);
return (s1+s1*s2)%mod;
}
else
{
s1=sum_mod(a,b/2-1);
s2=pow_mod(a,b/2);
return (s1+s2+a*s1*s2)%mod;
}
}
int main()
{
//freopen("out.txt","w",stdout);
makeprime();
int i,j;
scanf("%d%d",&n,&m);
if(n==0)
{printf("0\n");
return 0;
}
if(m==0||n==1)
{
printf("1\n");
return 0;
}
cnt=0;
dividedata();
int ans=1;
for(i=1;i<=cnt;i++)
ans=ans*sum_mod(p[i],k[i]*m)%mod;
printf("%d\n",ans);
return 0;
}