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 = ba�​=b. f(n)f(n) is the number of decomposition ways that n=abn=ab such that aaand 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

 

欧拉筛法学习。

代码：

#include <iostream>
#define MAX 10000000
using namespace std;
int t;
int n;
int num[MAX * 2 + 1],sum[MAX * 2 + 1];
bool vis[MAX * 2 + 1];
int prime[MAX + 1];
void init() {
    sum[1] = num[1] = 1;
    for(int i = 2;i <= MAX * 2;i ++) {
        if(!vis[i]) {
            prime[++ prime[0]] = i;
            num[i] = 2;
        }
        for(int j = 1;j <= prime[0] && i * prime[j] <= MAX * 2;j ++) {
            vis[i * prime[j]] = true;
            if(i % prime[j] == 0) {
                if(i / prime[j] % prime[j]) num[i * prime[j]] = num[i / prime[j]];
                break;
            }
            num[i * prime[j]] = num[i] * num[prime[j]];
        }
        sum[i] += sum[i - 1] + num[i];
    }
}
int main() {
    scanf("%d",&t);
    init();
    while(t --) {
        scanf("%d",&n);
        printf("%d
",sum[n]);
    }
    return 0;
}