计蒜客 30999.Sum-筛无平方因数的数 (ACM-ICPC 2018 南京赛区网络预赛 J)

J. Sum

• 26.87%
• 1000ms
• 512000K

 

A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integers could be decomposed into product of two square-free integers, there may be more than one decomposition ways. For example, 6 = 1\cdot 6=6 \cdot 1=2\cdot 3=3\cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=ab and n=ban=ba are considered different if a \not = ba̸=b. f(n)f(n) is the number of decomposition ways that n=abn=ab such that aa and bb are square-free integers. The problem is calculating \sum_{i = 1}^nf(i)∑i=1n​f(i).

Input

The first line contains an integer T(T\le 20)T(T≤20), denoting the number of test cases.

For each test case, there first line has a integer n(n \le 2\cdot 10^7)n(n≤2⋅107).

Output

For each test case, print the answer \sum_{i = 1}^n f(i)∑i=1n​f(i).

Hint

\sum_{i = 1}^8 f(i)=f(1)+ \cdots +f(8)∑i=18​f(i)=f(1)+⋯+f(8)
=1+2+2+1+2+4+2+0=14=1+2+2+1+2+4+2+0=14.

样例输入复制

2
5
8

样例输出复制

8
14

题目来源

ACM-ICPC 2018 南京赛区网络预赛

题意很好理解，就是求1-n中没有平方因数的数的个数。首先预处理出来，然后就直接输出就可以了。

类似欧拉函数筛素数，在筛素数的时候把平方因数筛掉就可以了。

代码:

 //J-筛无平方因数的数
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<bitset>
 #include<cassert>
 #include<cctype>
 #include<cmath>
 #include<cstdlib>
 #include<ctime>
 #include<deque>
 #include<iomanip>
 #include<list>
 #include<map>
 #include<queue>
 #include<set>
 #include<stack>
 #include<vector>
 using namespace std;
 typedef long long ll;

 const double PI=acos(-1.0);
 const double eps=1e-;
 const ll mod=1e9+;
 const int inf=0x3f3f3f3f;
 const int maxn=2e7+;
 const int maxm=+;
 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

 bool is_prime[maxn];
 int prime[maxn];
 ll f[maxn],ans[maxn];
 int h;

 void init(int n)
 {
     f[]=;
     for(int i=;i<n;i++){
         if(!is_prime[i]){
             prime[++h]=i;
             f[i]=;
         }
         for(int j=;j<=h&&i*prime[j]<n;j++){
             is_prime[i*prime[j]]=;
             if(i%prime[j]==){
                 if(i%(prime[j]*prime[j])==)
                     f[i*prime[j]]=;
                 else
                     f[i*prime[j]]=f[i]/;
                 break;
             }
             else{
                 f[i*prime[j]]=f[i]*;
             }
         }
     }
     for(int i=;i<maxn;i++)
         ans[i]=ans[i-]+f[i];
 }

 int main()
 {
     h=;
     init(maxn);
     int t;
     scanf("%d",&t);
     while(t--){
         int n;
         scanf("%d",&n);
         printf("%lld\n",ans[n]);
     }
     return ;
 }

溜了。。。