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 = 1cdot 6=6 cdot 1=2cdot 3=3cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=ab and n=ban=ba are considered different if a ot = 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(Tle 20)T(T≤20), denoting the number of test cases.

For each test case, there first line has a integer n(n le 2cdot 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-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=2e7+10;
const int maxm=100+10;
#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[1]=1;
    for(int i=2;i<n;i++){
        if(!is_prime[i]){
            prime[++h]=i;
            f[i]=2;
        }
        for(int j=1;j<=h&&i*prime[j]<n;j++){
            is_prime[i*prime[j]]=1;
            if(i%prime[j]==0){
                if(i%(prime[j]*prime[j])==0) 
                    f[i*prime[j]]=0;
                else 
                    f[i*prime[j]]=f[i]/2;
                break;
            }
            else{
                f[i*prime[j]]=f[i]*2;
            }
        }
    }
    for(int i=1;i<maxn;i++)
        ans[i]=ans[i-1]+f[i];
}

int main()
{
    h=0;
    init(maxn);
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        printf("%lld
",ans[n]);
    }
    return 0;
}

溜了。。。