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 = 50000000 + 10;
const int MOD = 9901;

inline void read(int &x)
{
    x = 0;char ch = getchar();char c = ch;
    while(ch > '9' || ch < '0')c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
    if(c == '-')x = -x;
}

int a,b,cnt,xishu[5010],zhishu[5010],ans;

int pow(int p, int k)
{
    int r = 1, base = p;
    for(;k;k >>= 1)
    {
        if(k & 1)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 == 0) return 1;
    if(k == 1) return (p + 1) % MOD;
    if(k & 1) return ((long long)(1 + pow(p, (k + 1)/2)) * (long long)(cal(p, (k - 1)/2))%MOD) % MOD;
    else return((long long)(pow(p, k/2) + 1) * (long long)(cal(p, k/2 - 1)) % MOD + pow(p, k)) % MOD;
}

int main()
{
    read(a),read(b);
    register int nn = sqrt(a) + 1;
    for(register int i = 2;i <= nn && a > 1;++ i)
        if(a % i == 0)
        {
            zhishu[++cnt] = i;
            while(a % i == 0) ++ xishu[cnt], a /= i;
        }
    if(a > 1)zhishu[++cnt] = a, xishu[cnt] = 1;
    ans = 1;
    for(register int i = 1;i <= cnt;++ i)
    {
        ans *= cal(zhishu[i], xishu[i] * b);
        ans %= MOD;
    } 
    printf("%d", ans%MOD);
    return 0;
}