SDUT 3257 Cube Number 简单数学

Posted 2020-07-12 shuguangzw

把所有数的立方因子除去，那么一个数可以和它组成立方的数是确定的，统计就行

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
const int N=1e6+5;
const int INF=0x3f3f3f3f;
int vis[N],prime[1005],cnt;
void getprime(){
  bool v[1020];
  memset(v,0,sizeof(v));
  for(int i=2;i*i<=1000;++i){
    if(v[i])continue;
    for(int j=i*i;j<=1000;j+=i)
      v[j]=1;
  }
  for(int i=2;i<=1000;++i)
    if(!v[i])prime[++cnt]=i;
}
int main(){
    int T,n;
    getprime();
    // printf("%d\\n",cnt);
    scanf("%d",&T);
    while(T--){
      memset(vis,0,sizeof(vis));
      scanf("%d",&n);
      LL ans=0;
      for(int i=1;i<=n;++i){
        LL t,t1;scanf("%lld",&t),t1=t,t=1;
        LL aim=1;bool flag=0;
        for(int j=1;j<=cnt&&prime[j]<=t1/prime[j];++j){
            int tmp=0;
            while(t1%prime[j]==0)t1/=prime[j],++tmp;
            if(tmp%3==1){
              t*=prime[j];
              aim*=prime[j];
              if(aim>N-5)flag=1;
              aim*=prime[j];
              if(aim>N-5)flag=1;
            }
            else if(tmp%3==2){
              t*=prime[j]*prime[j];
              aim*=prime[j];
              if(aim>N-5)flag=1;
            }
        }
        if(!flag){
          if(t1>1){
            t*=t1;
            aim*=t1;
            if(aim>N-5)flag=1;
            aim*=t1;
            if(aim>N-5)flag=1;
          }
          if(!flag)ans+=vis[aim];
        }
        ++vis[t];
      }
      printf("%lld\\n",ans);
    }
    return 0;
}

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