Ural2102：Michael and Cryptography(数论&素数)

The hacker Michael develops breakthrough password manager, which is called KEK (Keeper of Encrypted Keys). A distinctive feature of KEK is excellent security. To achieve this, Michael had to develop innovative encryption scheme. For example, in the well-known RSA scheme the sum of prime powers in the factorization is equal to 2, whereas in Michael’s scheme this sum is equal to 20!

However, the current version of the KEK runs very slow. Michael has found out that the problem is in the function of checking a modulus for correctness. This function should take the number n and answer, whether the sum of prime powers included in the factorization of n is equal to 20. Can you do this quickly?

Remember that the factorization of an integer is the representation of it in the form like p ₁ ^α₁ · p ₂ ^α₂ · ... · p _k ^α_k, where p _i are prime numbers, and α _i > 0. It is known that such representation is unique. Then the sum of powers looks likeα ₁ + α ₂ + ... + α _k.

Input

The only line contains an integer n (1 ≤ n ≤ 10 ¹⁸).

Output

If the sum of prime powers, included in the factorization of n, is equal to 20, then output “Yes”, otherwise output “No”.

Example

input	output
2	No
1048576	Yes
10000000000	Yes

题意：给定数字N(1e18级)，问将其唯一分解后（N=a1^p1*a2^p2...），幂的和（p1+p2+p3...）是否为20。

思路：根号N等于1e9级别，显然不能普通地分解因子来做。但是注意到20比较小，可以从20出发考虑：

           将N分解后不可能有两个大于1e6的因子。因为1e6*1e6*2^18>2e18。认识到这一点，说明最多只有一个大于1e6的因子。所以只要在[1,1e6]找19个素数因子，然后判断剩下的数是不是素数即可。

 

#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
const int maxn=2000000;
int p[maxn+10],vis[maxn+10],cnt;
void getprime()
{
    for(int i=2;i<=maxn;i++){
        if(!vis[i]) p[++cnt]=i;
        for(int j=1;p[j]*i<=maxn;j++){
            vis[i*p[j]]=1;
            if(i%p[j]==0) break;
        }
    }
}
bool isprime(ll x)
{
    for(int i=2;i*i<=x;i++)
      if(x%i==0) return false;
    return true;
}
int main()
{
    getprime();
    ll N; int num=0;
    scanf("%lld",&N);
    if(N<1024*1024){
        printf("No\n");
        return 0;
    }
    for(int i=1;i<=cnt;i++){
        while(N%p[i]==0){
            N/=p[i]; num++;
            if(num==20&&N==1){ printf("Yes\n"); return 0;}
            if(num>=20){ printf("No\n");return 0;}
        }
    }
    if(num<19) printf("No\n");
    else if(num==19&&N>1&&isprime(N)) printf("Yes\n");
    else printf("No\n");
    return 0;
}