POJ 1845 Sumdiv

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Sumdiv

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). 
運用不少知識點來自這點擊開啟連結

AC代碼如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define mod 9901#define M 10000#define ll long longusing namespace std;ll power(ll d,ll p)//冪次的最佳化{    ll ans=1;    while(p>0)    {        if(p%2)            ans=(ans*d)%mod;        p/=2;        d=(d*d)%mod;    }    return ans;}ll sum(ll d,ll p)//等比數列求和遞迴{    if(p==0)        return 1;    if(p==1)        return 1+d;    if(p%2==1)        return sum(d,p/2)*(1+power(d,p/2+1))%mod;    else return (sum(d,p/2-1)*(1+power(d,p/2))%mod+power(d,p))%mod;}int main(){    int i,j;    int a,b;    while(~scanf("%d%d",&a,&b))    {        int ds[M];        int po[M];        int tt=0;        for(i=2;i*i<=a;)//求a的因子        {            if(a%i==0)            {                ds[tt]=i;                po[tt]=0;                while(!(a%i))                {                    po[tt]++;                    a/=i;                }                tt++;            }            i==2?i++:i+=2;        }        if(a!=1)        {            ds[tt]=a;            po[tt++]=1;        }        ll ans = 1;        for(i=0;i<tt;i++)            ans=(ans*(ll)sum(ds[i],po[i]*b))%mod;        printf("%I64d\n",ans);    }}

POJ 1845 Sumdiv