[hdu-6623]Minimal Power of Prime

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6623

题意：一个质数p，找p质因子分解后，最小的 质因子次数

题解：

n先约去10^4内的质数因子，剩余为m，
剩余质数因子>10^4，则m最大的质数因子次数为4,几种可能为：m=p^4 m=p^2*q^2 m=p^3  m=p^2 ，除去前面几种m质因数分解，其他最小的质数因子次数都为1 :m=p*q*r*s,m=p^2*q*r,m=p^3*q...m=p^2*q,m=p*q,m=q

ps:在m开三次方的时候，用p=pow(m,1/3)有精度差，可以用差小于1e-6,或 二分(1~pow(m,1/3)+1)来找p。

官方题解：

Let's first factorize N using prime numbers not larger than N^1/5. And let's denote M as the left part, all the prime factors of M are larger than N^1/5. If M=1 then we are done, otherwise M can only be P², P³, P⁴ or P²*Q², here P and Q are prime numbers.

(1) If M^1/4 is an integer, we can know that M=P⁴. Update answer using 4 and return.

(2) If M^1/3 is an integer, we can know that M=P³. Update answer using 3 and return.

(3) If M^1/2 is an integer, we can know that M=P² or M=P²*Q². No matter which situation, we can always update answer using 2 and return.

(4) If (1)(2)(3) are false, we can know that answer=1.

Since there are just O(N^1/5/log(N)) prime numbers, so the expected running time is O(T*N^1/5/log(N)).

AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
ll prime[maxn],n,ans;
bool flag[maxn];
int cntprime;
void getl(){
    cntprime=0;
    for (ll i = 2; i <= 1e6; i++){
        if (!flag[i])  prime[++cntprime] = i;
        for (int j = 1; j <= cntprime && prime[j] * i <= 1e6; j++){
            flag[i * prime[j]] = true;
            if (i % prime[j] == 0)
            break;
        }
    }
}
void work(){
    ans=1e18;
    ll m=n,cnt,cnt1;
    for(int i=1;i<=cntprime&&prime[i]<10000&&prime[i]<=m;i++){
        if(m%prime[i]==0){
          cnt=0;
          while(m%prime[i]==0){cnt++;m/=prime[i];}
          ans=min(ans,cnt);
          if(ans==1) return ;
        }
    }
    
    if(m>1){
        cnt=sqrt(m);
        if(cnt*cnt==m){
            cnt1=sqrt(cnt);
            if(cnt1*cnt1==cnt){if(ans>4)ans=4;}
            else if(ans>2){ans=2;}
        }
        else {
            double q=pow(m,1.0/3.0);
            if(abs(q-ceil(q))<abs(q-floor(q)))cnt=ceil(q);
            else cnt=floor(q);
            if(cnt*cnt*cnt==m&&ans>3){ans=3;return ;}
            ans=1;return ;
        }
    }
}    
int main(){
    ll T;
    scanf("%lld",&T);
    getl();
    while(T --){
        scanf("%lld",&n);
        if(n<1e6&&!flag[n]){
            printf("1
");
            continue;
        }
        work();
        printf("%lld
",ans);
    }
    return 0;
}