POJ1845Sumdiv（求所有因子和 + 唯一分解定理）

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 17387		Accepted: 4374

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

题意：求A ^ B的所有因子的和；

分析：对A用唯一分解定理分解 A = p1 ^ a1 * p2 ^ a2 * p3 ^ a3 ... * pn ^ an

其中A的因子的个数为 （ 1 + a1) * ( 1 + a2 ) * ( 1 + a3 ) * ... * ( 1 + an)

则A的所有因子的和为 （1 + p1 + p1 ^ 2 + p1 ^ 3 ... + p1 ^ a1) * ( 1 + p2 ^ 1 + p2 ^ 2 + p3 ^ 3 + ... + p2 ^ a2 ) * ... * ( 1 + pn + pn ^ 2 + ... + pn ^ an)

求  a + a ^ 2 + a ^ 3 + ... + a ^ n 

如果n为奇数：  a + a ^ 2 + ... + a ^ (n / 2 )  + a ^ (n / 2 + 1) + ( a + a ^ 2 + ... + a ^ ( n / 2 ) ) *  a ^ ( n / 2 + 1）

如果n为偶数：  a + a ^ 2 + ... + a ^ (n / 2 ） + （ a + a ^ 2 + ... + a ^ ( n /. 2) ) * a ^ (n / 2) 

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 const int Max = ;
 const int Mod = ;
 int prime[Max + ],flag[Max],cnt;
 void get_prime()
 {
     cnt = ;
     memset(flag, , sizeof(flag));
     for(int i = ; i <= Max; i++)
     {
         if(flag[i] == )
         {
             flag[i] = ;
             prime[++cnt] = i;
             for(int j = i; j <= Max / i; j++)
                 flag[i * j] = ;
         }
     }
 }
 LL pow_mod(LL n, LL k)
 {
     LL res = ;
     while(k)
     {
         if(k & )
             res = res * n % Mod;
         n = n * n % Mod;
         k >>= ;
     }
     return res;
 }
 LL get_sum(LL n, LL m)
 {
     if(m == )
         return ;
     if(m & )
     {
         return get_sum(n, m / ) *(  + pow_mod(n, m /  + ) ) % Mod;
     }
     else
     {
         return ( get_sum(n, m /  - ) * ( + pow_mod(n, m /  + )) % Mod + pow_mod(n, m / ) ) % Mod;
     }
 }
 int main()
 {
     LL a,b;
     get_prime();
     while(scanf("%I64d%I64d", &a, &b) != EOF)
     {
         LL ans = ;
         if(a ==  && b) //特殊情况
             ans = ;
         LL m;
         for(int i = ; i <= cnt; i++)
         {
             if(prime[i] > a)
                 break;
             m = ;
             if(a % prime[i] == )
             {
                 while(a % prime[i] == )
                 {
                     a = a / prime[i];
                     m++;
                 }
                 m = m * b;  // m要设成LL，否则这里会溢出
                 ans = ans * get_sum((LL)prime[i], m) % Mod;
             }
         }
         if(a > )
             ans = ans * get_sum(a, b) % Mod;
         printf("%I64d\n", ans);
     }
     return ;
 }