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POJ 1845 Sumdiv

2024-07-13 06:42:01   210人阅读

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). 

运用不少知识点来自这点击打开链接

AC代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define mod 9901
#define M 10000
#define ll long long
using namespace std;

ll power(ll d,ll p)//幂次的优化
{
    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)//等比数列求和递归
{
    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


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