[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 N1/5. And let's denote M as the left part, all the prime factors of M are larger than N1/5. 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 M1/4 is an integer, we can know that M=P4. Update answer using 4 and return.

(2) If M1/3 is an integer, we can know that M=P3. Update answer using 3 and return.

(3) If M1/2 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(N1/5/log(N)) prime numbers, so the expected running time is O(T*N1/5/log(N)).

AC代码

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn=1e6+10;
 5 ll prime[maxn],n,ans;
 6 bool flag[maxn];
 7 int cntprime;
 8 void getl(){
 9     cntprime=0;
10     for (ll i = 2; i <= 1e6; i++){
11         if (!flag[i])  prime[++cntprime] = i;
12         for (int j = 1; j <= cntprime && prime[j] * i <= 1e6; j++){
13             flag[i * prime[j]] = true;
14             if (i % prime[j] == 0)
15             break;
16         }
17     }
18 }
19 void work(){
20     ans=1e18;
21     ll m=n,cnt,cnt1;
22     for(int i=1;i<=cntprime&&prime[i]<10000&&prime[i]<=m;i++){
23         if(m%prime[i]==0){
24           cnt=0;
25           while(m%prime[i]==0){cnt++;m/=prime[i];}
26           ans=min(ans,cnt);
27           if(ans==1) return ;
28         }
29     }
30     
31     if(m>1){
32         cnt=sqrt(m);
33         if(cnt*cnt==m){
34             cnt1=sqrt(cnt);
35             if(cnt1*cnt1==cnt){if(ans>4)ans=4;}
36             else if(ans>2){ans=2;}
37         }
38         else {
39             double q=pow(m,1.0/3.0);
40             if(abs(q-ceil(q))<abs(q-floor(q)))cnt=ceil(q);
41             else cnt=floor(q);
42             if(cnt*cnt*cnt==m&&ans>3){ans=3;return ;}
43             ans=1;return ;
44         }
45     }
46 }    
47 int main(){
48     ll T;
49     scanf("%lld",&T);
50     getl();
51     while(T --){
52         scanf("%lld",&n);
53         if(n<1e6&&!flag[n]){
54             printf("1\n");
55             continue;
56         }
57         work();
58         printf("%lld\n",ans);
59     }
60     return 0;
61 }

 

posted @ 2019-08-01 10:28  conver^_^  阅读(213)  评论(0)  编辑  收藏  举报