POJ 1845 Sumdiv

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10515		Accepted: 2477

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

求 A^B 的约数和 mod 9901

把A因式分解成 p1^a1*p2^a2*..*pk^ak
然后用约数求和函数

这里关键问题是 有可能出现 gcd(pi-1，9901)!=1
这样的话就不能求逆元了、、、、 

如果没有逆元
那么 对于  1+p1+p1^2...p^n

作以下处理 
n%2==1
 原式=(1+p+p^2+..p^(n/2))(1+p^(n/2+1))
n%2==0
n=n-1;
 原式=(1+p+p^2+..p^(n/2))(1+p^(n/2+1))+p^(n+1);

二分递归求解

我忘记考虑 a是9901的倍数情况了，那样 tp会为负，Wa的我好痛苦、、、

#include <iostream>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <algorithm>

using namespace std;

int x,y;

#define Mod 9901

void EGcd(int a,int b)

{

    if(b==0)

    {

        x=1;

        y=0;

        return ;

    }

    EGcd(b,a%b);

    int x0=y;

      y=x-a/b*y;

      x=x0;

}

int gcd(int a,int b)

{

    int r;

    while(r=a%b){a=b;b=r;}

    return b;

}

#define N 10002

bool h[N];

int pm[N>>1];

void Getprime()

{

    int cnt=1;

    int i,j;

    pm[0]=2;

    for(i=2;i<N;i+=2)

     h[i]=1;

    for(i=3;i<N;i+=2)

     if(!h[i])

     {

         pm[cnt++]=i;

         for(j=i*i;j<N;j+=i)

          h[j]=1;

     }

}

int Pw(int a,int b)

{

    int t;

    for(t=1;b>0;b=(b>>1),a=(a%Mod)*(a%Mod)%Mod)

        if(b&1) t=(t*a)%Mod;

    return t;

}

int bn(int a,int n)

{

    if(n==2)

    {

        return (1+a+(a%Mod)*(a%Mod)%Mod)%Mod;

    }

    if(n==1)

    {

        return (1+a)%Mod;

    }

    if(n%2)

    {

        return bn(a,n/2)*(1+Pw(a,n/2+1)%Mod)%Mod;

    }

       n--;

        return bn(a,n/2)*(1+Pw(a,n/2+1)%Mod)%Mod+Pw(a,n+1);

}

int rc[12][2];

int main()

{

    Getprime();

    int a,b;

    int tp;

    while(scanf("%d %d",&a,&b)!=EOF)

    {

        if(a==0){printf("0\n");continue;}

        if(b==0||a==1){printf("1\n");continue;}

        tp=1;

        int m=(int)sqrt(a*1.0);

        int i,cnt=0;

        for(i=0;pm[i]<=m;i++)

         if(a%pm[i]==0)

         {

             rc[cnt][0]=pm[i];

             rc[cnt][1]=0;

             while(a%pm[i]==0){a/=pm[i];rc[cnt][1]++;}

             cnt++;

         }

         if(a>1) { rc[cnt][0]=a;rc[cnt++][1]=1;}



         for(i=0;i<cnt;i++)

         {



             if(gcd(rc[i][0]-1,Mod)==1)

             {

                 tp=tp*(Pw(rc[i][0],rc[i][1]*b+1)-1)%Mod;

                 while(tp<0)tp+=9901;//忘记加这个、、、Wa好久

                 EGcd(rc[i][0]-1,9901);

                 while(x<0)x+=9901;

                 tp=tp*(x%9901)%9901;

             }

             else

             { 

                 tp=tp*bn(rc[i][0],rc[i][1]*b)%9901;

             }

         }

        printf("%d\n",tp);

    }





    return 0;

}