2020牛客寒假算法基础集训营6 E.立方数（唯一分解定理 素数筛）

https://ac.nowcoder.com/acm/contest/3007/E

 

放下题解

 

 

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 32640;
 5 ll vis[maxn*100],prime[maxn],prime_3[maxn];
 6 ll tot = 0;
 7 void getPrime(){
 8     memset(vis,0,sizeof(vis));
 9     for(ll i=2;i<=maxn;i++){
10         if(!vis[i])prime[tot++]=i,prime_3[tot-1] = i*i*i;
11         for(ll j=0;j<tot;j++){
12             if(prime[j]*i>maxn)break;
13             vis[prime[j]*i]=1;
14             if(i%prime[j]==0)break;
15         }
16     }
17 }
18 
19 int main(){
20     getPrime();
21     int t;scanf("%d",&t);
22     while(t--){
23         ll n;
24         ll ans = 1;
25         scanf("%lld",&n);
26         for(int i = 0;i<tot&&prime[i]<=n;i++){
27             while(n%prime_3[i] == 0){
28                 ans *=prime[i];
29                 n/=prime_3[i];
30             }
31             while(n%prime[i] == 0){
32                 n/=prime[i];
33             }
34         }
35         int L = 1, R = 1000000;
36         while(L<=R){//二分试除
37             int mid = (L+R)/2;
38             if((ll)mid*mid*mid<n) L = mid + 1;
39             else R = mid - 1;
40         }
41         if((ll)L*L*L == n) ans*=L;
42         printf("%lld\n",ans);
43     }
44     return 0;
45 }

 

posted @ 2020-02-18 16:10  AaronChang  阅读(214)  评论(0)    收藏  举报