POJ 1845 Sumdiv (大数据因子和)
Sumdiv
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 17287 | Accepted: 4343 |
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).
题意：求出A^B的因子和
思路：根据唯一分解定理得A=p1^k1*p2^k2*...*pn^kn,所以说A^B=p1^(k1*B)*p2^(k2*B)*...*pn^(kn*B),所以说根据因子和公式得(1+p1+p1^2+...+p1^(k1*B))*(1+p2+p2^2+...+p2^(k2*B))*...*(1+pn+pn^2+...+pn^(kn*B)),结果就是一个个等比数列的乘积，但是数据是5e8，这样直接按照等比数列的求和公式写会超时或者RE，所以说要优化等比数列求和，用折半二分递归的方法求等比数列的和，这里就先不说了，等后来总结的时候再深究。
ac代码：
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 100010
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-10
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
struct s
{
int num;
int cnt;
}p[MAXN];
int fun(int a,int n,int MOD)
{
if(n==0)
return 1;
if(n%2)
return ((1+powmod(a,n/2+1,MOD))%MOD*fun(a,n/2,MOD)%MOD)%MOD;
else
return (powmod(a,n/2,MOD)+(1+powmod(a,n/2+1,MOD))%MOD*fun(a,(n-1)/2,MOD)%MOD)%MOD;
}
int main()
{
int a,b,i;
while(scanf("%d%d",&a,&b)!=EOF)
{
int x=a;
int ccnt=0;
for(i=2;i*i<=x;i++)
{
if(x%i==0)
{
p[ccnt].num=i;
int t=0;
while(x%i==0)
x/=i,t++;
p[ccnt++].cnt=t;
}
if(x==1)
break;
}
if(x!=1)
{
p[ccnt].num=x;
p[ccnt].cnt=1;
ccnt++;
}
LL ans=1;
for(i=0;i<ccnt;i++)
ans=(ans*fun(p[i].num,p[i].cnt*b,9901))%9901;
printf("%d\n",ans);
}
return 0;
}