ACM-ICPC 2018 南京赛区网络预赛 J.Sum

Sum

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：定义f[i]函数代表i=a*b的对数，其中a和b都不能是平方数的倍数，a*b与b*a不相同，t组样例，给出n，求1~n的f[i]之和

 类似于素数筛＋前缀和

 

#include <bits/stdc++.h>
#define ll long long int
#define N 20000002
using namespace std;
int sum[N];
bool an[N];
int bn[N];
int cnt = 0;
void init(){
    an[0] = true;
    for(int i=2;i*i<N;i++){
        ll k = i*i;
        for(int j = k;j<N;j+=k){
            an[j] = true;
        }
    }
    for(int i=1;i<N;i++){
        if(!an[i]){
            sum[i] = sum[i-1]+1;
            bn[cnt++] = i;
        }else
            sum[i] = sum[i-1];
    }
}
int t,n;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    init();
    cin>>t;
    while(t--){
        cin>>n;
        ll ans = 0;
        for(int i=0;i<cnt&&bn[i]<=n;i++){
            int pox = n/bn[i];
            ans += sum[pox];
        }
        cout<<ans<<endl;
    }
    return 0;
}