POJ 1845 Sumdiv(數論+快速冪)

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 19576		Accepted: 4930

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

首先因數和定理p^k=(1+p^1+p^2+...p^k)；

之後構造矩陣求p^k%MOD

之前想過利用逆元來做，但是當(p-1)%MOD==0時不存在逆元，所以就沒有繼續想。但是逆元的思想比較直接簡便。

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<math.h>
using namespace std;
#define ll long long
#define F(x,a,b) for(ll x=a;x<=b;x++)
#define me(x) memset(x,0,sizeof(x))
#define MOD 9901
#define _fast F(i,1,3)F(j,1,3)F(k,1,3) b[i][j]=(b[i][j]+a[i][k]*a[k][j])%MOD;
#define _reset F(i,1,3)F(j,1,3){a[i][j]=b[i][j]; b[i][j]=0;}
#define _orz F(i,1,3)F(j,1,3)F(k,1,3) b[i][j]=(b[i][j]+a[i][k]*c[k][j])%MOD;
ll p[100005],isp[100005],cnt,x,y;
ll b[10][10],a[10][10],c[10][10];
struct p{ll p,k;}rec[100000];
void _pr(ll x)
{
    me(isp);me(p);
    F(i,2,x)
    {
      if (!p[i]) isp[++p[0]]=i;
      F(j,1,p[0])
      {

          if (isp[j]*i>x) break;
          p[isp[j]*i]=1;
          if (i%isp[j]==0) break;
      }
    }
}
void _s(ll x)
{
    cnt=0;
    me(rec);
    int t=x;
    F(i,1,p[0])
    {
        if (isp[i]*isp[i]>x) break;
        if (t%isp[i]==0)
        {
           rec[++cnt].p=isp[i];
           while (t%isp[i]==0)
           {
               t/=isp[i];
               rec[cnt].k++;
           }
        }
    }
    if (t>1) {rec[++cnt].p=t;rec[cnt].k=1;}
}
void _f(ll kk)
{
    if (kk==1) return;
    if (kk&1){_f(kk-1); _orz _reset}
    else {_f(kk/2); _fast _reset}
}
int main()
{
    ll n,k;
    _pr(100000);
    cin>>n>>k;
    if (k==0) {cout<<"1"<<endl;return 0;}
    cnt=0;
    _s(n);
    ll ans=1; a[1][1]=1;a[1][3]=1;a[2][1]=1;
    F(i,1,cnt)
    {
        me(a);me(b);me(c);
        a[1][1]=1;a[1][3]=1;a[2][1]=1;
        a[3][3]=rec[i].p;
        F(i,1,3)F(j,1,3) c[i][j]=a[i][j];
        _f(rec[i].k*k);
        ans=ans*(a[1][1]*1+a[1][3]*rec[i].p)%MOD;
    }
    cout<<ans<<endl;
}