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).
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).
两个重要的结论:
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)
用递归二分求等比数列1+pi+pi^2+pi^3+...+pi^n:
一、若n为奇数,一共有偶数项,则:
1 + p + p^2 + p^3 +...+ p^n
= (1+p^(n/2+1)) + p * (1+p^(n/2+1)) +...+ p^(n/2) * (1+p^(n/2+1))
这里将其分成两半,提后一半的公因式p^(n/2+1)
= (1 + p + p^2 +...+ p^(n/2)) * (1 + p^(n/2+1))
二、若n为偶数,一共有奇数项,则:
1 + p + p^2 + p^3 +...+ p^n
= (1+p^(n/2+1)) + p * (1+p^(n/2+1)) +...+ p^(n/2-1) * (1+p^(n/2+1)) + p^(n/2)
可知n-1为偶数,对前偶数项进行上述操作再加上最后一项p^n
= (1 + p + p^2 +...+ p^(n/2-1)) * (1+p^(n/2)) + p^n;
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); } }