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.
樣例輸入複製
2 5 8
樣例輸出複製
8 14
題目來源
思路: 相同因子個數不超過2的數稱爲squarefree(SF),F【i】= A*B的種數,AB都是SF,AB不相等的時候AB 和BA算兩種方案,求1~n內F【i】前綴和。
魔改線性篩,太6了,Orz..................
代碼:
#include<bits/stdc++.h>
using namespace std;
const int maxn=2e7+5;
long long ans[maxn];
int tot,prime[maxn],vis[maxn],f[maxn];///prime保存遍歷到的素數
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<maxn;i++){
if(!vis[i]){
prime[tot++]=i;
f[i]=2;
}
for(int j=0;j<tot&&i*prime[j]<maxn;j++){
vis[i*prime[j]]=1;
if(i%prime[j]==0){
if(i%(prime[j]*prime[j])==0) f[i*prime[j]]=0;///大於等於三次方GG
else f[i*prime[j]]=f[i]/2;///等於二次方得變一半,相當在原基礎上少了另一個素數
break;
}
else f[prime[j]*i]=f[i]*2;
}
ans[i]=ans[i-1]+f[i];
}
}
int main(){
doit();
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
printf("%lld\n",ans[n]);
}
}