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 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

 

思路: 相同因子个数不超过2的数称为squarefree（SF），F【i】= A*B的种数，AB都是SF，AB不相等的时候AB 和BA算两种方案,求1~n内F【i】前缀和。

魔改线性筛，太6了，Orz..................

 

代码：

#include
using namespace std;
const int  maxn=2e7+5;
long long ans[maxn];
int tot,prime[maxn],vis[maxn],f[maxn];///prime保存遍历到的素数

void doit(){
    memset(vis,0,sizeof(vis));
    memset(prime,0,sizeof(prime));
    memset(f,0,sizeof(f));

    prime[1]=1;
    ans[1]=1;
    tot=0;

    for(int i=2;i