POJ1485 Sumdiv

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 22680		Accepted: 5660

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

 

【题目大意】

　　求A^B的所有约数的和

【题解】

　　把A唯一分解，不难得到答案：

 　　(1+p1+p1^2+……+p1^(φ1*B)) × (1+p2+p2^2+

　　……+p2^(φ2*B)) ×……× (1+pn+pn^2+……+pn^(φn*B))

　　问题转化为等比数列求和，此处MOD为质数，存在逆元，但对于更一般的情况，采用分治法，复杂度多一个Log

　　cal(p,k) = p^0 + p^1 + p^2 + ... + p^k

　　if k&1

　　　　cal(p,k) = (1 + p^((k + 1)/2))*cal(p, (k-1)/2)

　　else

　　　　cal(p,k) = (1 + p^(k/2)) * cal(p, k/2 - 1) + p^k

　　快速幂即可

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cstdlib>
 #include <cmath>
 
 const int INF = 0x3f3f3f3f;
 const int MAXN =  + ;
 const int MOD = ;
 
 inline void read(int &x)
 {
     x = ;char ch = getchar();char c = ch;
     while(ch > '' || ch < '')c = ch, ch = getchar();
     while(ch <= '' && ch >= '')x = x *  + ch - '', ch = getchar();
     if(c == '-')x = -x;
 }
 
 int a,b,cnt,xishu[],zhishu[],ans;
 
 int pow(int p, int k)
 {
     int r = , base = p;
     for(;k;k >>= )
     {
         if(k & )r = ((long long)r * base) % MOD;
         base = ((long long)base * base % MOD);
     }
     return r % MOD;
 }
 
 //求解1 + p + p^2 + p^3 + ... + p^k
 int cal(int p, int k)
 {
     if(k == ) return ;
     if(k == ) return (p + ) % MOD;
     if(k & ) return ((long long)( + pow(p, (k + )/)) * (long long)(cal(p, (k - )/))%MOD) % MOD;
     else return((long long)(pow(p, k/) + ) * (long long)(cal(p, k/ - )) % MOD + pow(p, k)) % MOD;
 }
 
 int main()
 {
     read(a),read(b);
     register int nn = sqrt(a) + ;
     for(register int i = ;i <= nn && a > ;++ i)
         if(a % i == )
         {
             zhishu[++cnt] = i;
             while(a % i == ) ++ xishu[cnt], a /= i;
         }
     if(a > )zhishu[++cnt] = a, xishu[cnt] = ;
     ans = ;
     for(register int i = ;i <= cnt;++ i)
     {
         ans *= cal(zhishu[i], xishu[i] * b);
         ans %= MOD;
     }
     printf("%d", ans%MOD);
     return ;
 }

POJ1848 Sumdiv