POJ1845——Sumdiv

文章转自：http://blog.csdn.net/a15110103117

Sumdiv

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 19376 Accepted: 4876

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：

给出a，b(0<=a,b<=50000000)，求出q为a，n为b的等比数列和。例如：1+2+2^2+2^3+......

• 

思路：

之前我还以为直接上公式再取余就完了，之后就惨不忍睹。

真正的做法是：

1.先把a分解成小数，比如36^100=2^200*3^200（至于为什么一定要分解就不清楚了，我直接递归求和是WA）

2.递归求和

n为奇数时：(1+p+p^2...p^(n/2))*(1+p^(n/2+1))

n为偶数时：(1+p+p^2...p^(n/2-1))*(1+p^(n/2+1))+p^(n/2)

这个式子可以在纸上推出来，黑色部分做递归就可以了，当n=0时就是递归的终点。

对于乘方需要快速幂，这里不给予叙述，详情请咨询百度。

#include <iostream>#include <cstdlib>#include <cstdio>using namespace std;long long sort_pow(int a, int b)                 //二分求幂{    long long r = 1, base = a;    while( b )    {        if( b%2==1 )            r = ( r*base )%9901;        base = ( base*base )%9901;        b /= 2;    }    return r;}long long multi(int a, long long b)       //递归求和{    if( b==0 )        return 1;    else if( b%2==1 )        return multi(a,b/2)*(1+sort_pow(a,b/2+1))%9901;    else        return (multi(a,b/2-1)*(1+sort_pow(a,b/2+1))%9901 + sort_pow(a,b/2))%9901;}long long dell( int a, int b ){    long long mul = 1;    int cnt;    for( int i = 2; i*i <= a;i++ )       //遍历到  sqrt(a)  就好，类似原始的素数求法    {        cnt = 0;        while( a%i==0 )      //因式分解        {            a /= i;            cnt++;        }        mul = mul * multi(i,b*cnt)%9901;    }    if( a!=1 )    {        mul = mul*multi(a,b)%9901;    }    return mul;}int main(){    int a, b;    cin>>a>>b;    cout<<dell(a,b)<<endl;    return 0;}