A^B约数和

转自https://www.cnblogs.com/aininot260/p/9574789.html

POJ1845

首先把A写成唯一分解定理的形式

分解时让A对所有质数从小到大取模就好了

然后就有：A = p1^k1 * p2^k2 * p3^k3 *...* pn^kn

然后有： A^B = p1^(k1*B) * p2^(k2*B) *...* pn^(kn*B);

约数和公式：

对于已经分解的整数A=(p1^k1)*(p2^k2)*(p3^k3)*....*(pn^kn)

有A的所有因子之和为

S = (1+p1+p1^2+p1^3+...p1^k1) * (1+p2+p2^2+p2^3+….p2^k2) * (1+p3+ p3^3+…+ p3^k3) * .... * (1+pn+pn^2+pn^3+...pn^kn)

那么A^B就可以是

sum = [1+p1+p1^2+...+p1^(a1*B)] * [1+p2+p2^2+...+p2^(a2*B)] *...* [1+pn+pn^2+...+pn^(an*B)].

求等比数列1+pi+pi^2+pi^3+...+pi^n

（1）若n为奇数,一共有偶数项，则：
      1 + p + p^2 + p^3 +...+ p^n
  </span>= (<span style="color: #800080;">1</span>+p^(n/<span style="color: #800080;">2</span>+<span style="color: #800080;">1</span>)) + p * (<span style="color: #800080;">1</span>+p^(n/<span style="color: #800080;">2</span>+<span style="color: #800080;">1</span>)) +...+ p^(n/<span style="color: #800080;">2</span>) * (<span style="color: #800080;">1</span>+p^(n/<span style="color: #800080;">2</span>+<span style="color: #800080;">1</span><span style="color: #000000;">))
  </span>= (<span style="color: #800080;">1</span> + p + p^<span style="color: #800080;">2</span> +...+ p^(n/<span style="color: #800080;">2</span>)) * (<span style="color: #800080;">1</span> + p^(n/<span style="color: #800080;">2</span>+<span style="color: #800080;">1</span><span style="color: #000000;">))

上式红色加粗的前半部分恰好就是原式的一半，那么只需要不断递归二分求和就可以了，后半部分为幂次式，将在下面第4点讲述计算方法。

（2）若n为偶数,一共有奇数项,则:
1 + p + p^2 + p^3 +…+ p^n

  </span>= (<span style="color: #800080;">1</span>+p^(n/<span style="color: #800080;">2</span>+<span style="color: #800080;">1</span>)) + p * (<span style="color: #800080;">1</span>+p^(n/<span style="color: #800080;">2</span>+<span style="color: #800080;">1</span>)) +...+ p^(n/<span style="color: #800080;">2</span>-<span style="color: #800080;">1</span>) * (<span style="color: #800080;">1</span>+p^(n/<span style="color: #800080;">2</span>+<span style="color: #800080;">1</span>)) + p^(n/<span style="color: #800080;">2</span><span style="color: #000000;">)
  </span>= (<span style="color: #800080;">1</span> + p + p^<span style="color: #800080;">2</span> +...+ p^(n/<span style="color: #800080;">2</span>-<span style="color: #800080;">1</span>)) * (<span style="color: #800080;">1</span>+p^(n/<span style="color: #800080;">2</span>+<span style="color: #800080;">1</span>)) + p^(n/<span style="color: #800080;">2</span><span style="color: #000000;">);

上式红色加粗的前半部分恰好就是原式的一半，依然递归求解

p^n直接快速幂就可以了

 1 #include<cstdio>
 2 #include<cstring>
 3 const int mod=9901;
 4 const int maxn=10005;
 5 int A,B;
 6 int fatcnt;
 7 int prime[maxn];
 8 long long factor[10][2];
 9 void get_prime()
10 {
11     memset(prime,0,sizeof(prime));
12     for(int i=2;i<=maxn;i++)
13     {
14         if(!prime[i]) prime[++prime[0]]=i;
15         for(int j=1;j<=prime[0]&&prime[j]<=maxn/i;j++)
16         {
17             prime[prime[j]*i]=1;
18             if(i%prime[j]==0) break;
19         }
20     }
21 }
22 int get_factors(long long x)
23 {
24     fatcnt=0;
25     long long tmp=x;
26     for(int i=1;prime[i]<=tmp/prime[i];i++)
27     {
28         factor[fatcnt][1]=0;
29         if(tmp%prime[i]==0)
30         {
31             factor[fatcnt][0]=prime[i];
32             while(tmp%prime[i]==0)
33             {
34                 factor[fatcnt][1]++;
35                 tmp/=prime[i];
36             }
37             fatcnt++;
38         }
39     }
40     if(tmp!=1)
41     {
42         factor[fatcnt][0]=tmp;
43         factor[fatcnt++][1]=1;
44     }
45     return fatcnt;
46 }
47 long long pow_mod(long long a,long long n)
48 {
49     long long res=1;
50     long long tmp=a%mod;
51     while(n)
52     {
53         if(n&1)
54         {
55             res*=tmp;
56             res%=mod;
57         }
58         n>>=1;
59         tmp*=tmp;
60         tmp%=mod;
61     }
62     return res;
63 }
64 long long sum(long long p,long long n)
65 {
66     //1+p+p^2+````+p^n
67     if(p==0) return 0;
68     if(n==0) return 1;
69     if(n&1)
70         return ((1+pow_mod(p,n/2+1))%mod*sum(p,n/2)%mod)%mod;
71     else
72         return ((1+pow_mod(p,n/2+1))%mod*sum(p,n/2-1)+pow_mod(p,n/2)%mod)%mod;
73 }
74 int main()
75 {
76     get_prime();
77     while(scanf("%d%d",&A,&B)==2)
78     {
79         get_factors(A);
80         long long ans=1;
81         for(int i=0;i<fatcnt;i++)
82         {
83             ans*=(sum(factor[i][0],B*factor[i][1])%mod);
84             ans%=mod;
85         }
86         printf("%I64d\n",ans);
87     }
88     return 0;
89 }