ACM-ICPC 2018 南京赛区网络预赛 J. Sum

A square-free integer is an integer which is indivisible by any square number except 1. For example, 6 = 2 * 3,6=2*3 is square-free, but 12 = 2^2 * 3,12=2^2*3 is not,

because 2^2 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* 6, 6=2* 3=3*2   6=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≤20), denoting the number of test cases.

For each test case, there first line has a integer 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

]

 

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
#include <set>
#include <queue>
#include <cmath>
#define  ll long long 
#define  M 20000004
#define  P  pair<int,int>
using namespace std;
bool  vis[M];
int sum[M];
void  init()
{
    int u=sqrt(M+0.5);
    for(int i=1;i<=M;i++) vis[i]=1;
    for(int i=2;i<=u;i++)
    {
        int k=i*i;
        for(int j=k;j<M;j+=k)
        {
            if(vis[j])  vis[j]=0;
        }
    }
    /*
    1 2 3 4 5 6 7 8
    1 2 3 3 4 5 6 6
    */
    for(int i=1;i<M;i++)
    {
        if(vis[i]){
            sum[i]=sum[i-1]+1;
        }
        else  sum[i]=sum[i-1];
    }
}
int t,n;
int  main()
{
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int u=sqrt(n+0.5);
        ll ans=0;
        /*
        n==8
        i==1 : 1*8最大是8.
        (1,2),(1,3),(1,5),(1,6),(1,7) 
        (2,1),(3,1),(5,1),(6,1),(7,1)
        (1,1) 
        那么 2*5+1==11
        i==2  : 2*4最大是 4
        (2,3) 
        (3,2)
        (2,2)
        1*2+1==3
        11+3==14
        */
        for(int i=1;i<=u;i++){
            if(!vis[i])  continue;
            int y=n/i;
            ans+=(ll)(sum[y]-sum[i])*2+1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}