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=1nf(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=1nf(i).
Hint
\sum_{i = 1}^8 f(i)=f(1)+ \cdots +f(8)∑i=18f(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
2 5 8
样例输出
8 14
题意：找出一个数的所有乘数组合，使这些乘数组合中的任一因数都不存在完全平方因数。求从1到n的的所有数的乘数组合之和。
思路：用线性筛法，求出所有数的最小质因数，不难发现，假如一个数的因数中有一质因数的三次方，则其定然不存在符合上述条件 的组合，而假使存在一质因数的平方，那么这个质因数的平方就必须分配到一个组合的两侧，因此其组合数的数量就等于该数除以这一质因数的数的组合数，而如果只存在一个质因数时，则等于该数的组合数除去该质因数乘2.
AC代码：
#include<cstdio>
using namespace std;
typedef long long LL;
const int size=2e7+10;
int minPrime[size];
int ans[size];
bool prime[size];
int p[size];
int tot;
void init()
{
int n=size;
prime[1]=true;
for(int i=1;i<=n;i++) prime[i]=true;
for(int i=2;i<n;i++)
{
if(prime[i] ) p[tot++]=i,minPrime[i]=i;
for(int j=0;j<tot&&i*p[j]<n;j++)
{
prime[i*p[j]]=false;
minPrime[i*p[j]]=p[j];
if(i%p[j]==0) break;
}
}
ans[1]=1;
for(int i=2;i<size;i++)
{
int Mip=minPrime[i];
if(i>=Mip*Mip*Mip&&i%(Mip*Mip*Mip)==0) ans[i]=0;
else if(i>=Mip*Mip&&i%(Mip*Mip)==0)
{
ans[i]=ans[i/(Mip*Mip)];
}
else
{
ans[i]=ans[i/Mip]*2;
}
}
}
int main()
{
init();
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
LL res=0;
for(int i=1;i<=n;i++)
{
res=res+ans[i];
}
printf("%lld\n",res);
}
}