poj_1845_Sumdiv

Posted 2020-10-15 会飞的雅蠛蝶

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).

 

 

 

这道题目令我很不解的就是在等比求和的时候。大佬们分奇偶项数用二分递归求。而我的思路是乘法逆元*P₁(P₁^n  -1)，也就是等比数列求和，在我出的数据中所得结果相同（rand随机出数)。。。不知道是什么原理

 

 

一开始想到的是A%mod^B，一举大数据直接炸。很快就想到唯一分解定理，然后等比求和，再然后就晾在等比求和上了。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include "cstdio"
using namespace std;
#define ll long long
#define mod 9901
#define N 1000010
ll prime[N];
bool vis[N];
ll p[N];
ll pn=0;
ll POW(ll a,ll n)
{
    ll base=a,ret=1;
    while(n)
    {
        if(n&1) ret=(ret%mod*base)%mod;
        base=(base*base)%mod;
        n>>=1;
    }
    return ret%mod;
}
__int64 sum(__int64 p,__int64 n)  //递归二分求 (1 + p + p^2 + p^3 +...+ p^n)%mod
{                          //奇数二分式 (1 + p + p^2 +...+ p^(n/2)) * (1 + p^(n/2+1))
    if(n==0)               //偶数二分式 (1 + p + p^2 +...+ p^(n/2-1)) * (1+p^(n/2+1)) + p^(n/2)
        return 1;
    if(n%2)  //n为奇数,
        return (sum(p,n/2)*(1+POW(p,n/2+1)))%mod;
    else     //n为偶数
        return (sum(p,n/2-1)*(1+POW(p,n/2+1))+POW(p,n/2))%mod;
}
int main()
{
    for (int i = 2; i < N; i++) {
        if (vis[i]) continue;
        prime[pn++] = i;
        for (int j = i; j < N; j += i)
            vis[j] = 1;
    }
    ll a,b;
    while(~scanf("%lld%lld",&a,&b))
    {
        memset(p,0,sizeof(p));
        ll ans=1;
        for(int i=0;prime[i]*prime[i]<=a;i++)
        {
            ll tem=0;
            while(a%prime[i]==0)
            {
                tem++;
                a/=prime[i];
            }
            if(tem)
            {
                p[prime[i]]=tem;
            }
        }
        if(a!=1)
        {
                ll rev=POW(a-1,9899);
                ll res=sum(a,b);
                ans=ans*res%mod;
        }
        for(int i=0;i<pn;i++)
        {

            if(p[prime[i]])
            {
                ll rev=POW(prime[i]-1,9899);
                ll res=sum(prime[i],p[prime[i]]*b);
                ans=ans*res%mod;
            }
        }
        cout<<ans<<endl;
    }
}