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 = ba?=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
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
1 #include <bits/stdc++.h>
2 #define ll long long int
3 #define N 20000002
4 using namespace std;
5 int sum[N];
6 bool an[N];
7 int bn[N];
8 int cnt = 0;
9 void init(){
10 an[0] = true;
11 for(int i=2;i*i<N;i++){
12 ll k = i*i;
13 for(int j = k;j<N;j+=k){
14 an[j] = true;
15 }
16 }
17 for(int i=1;i<N;i++){
18 if(!an[i]){
19 sum[i] = sum[i-1]+1;
20 bn[cnt++] = i;
21 }else
22 sum[i] = sum[i-1];
23 }
24 }
25 int t,n;
26 int main(){
27 ios::sync_with_stdio(false);
28 cin.tie(0),cout.tie(0);
29 init();
30 cin>>t;
31 while(t--){
32 cin>>n;
33 ll ans = 0;
34 for(int i=0;i<cnt&&bn[i]<=n;i++){
35 int pox = n/bn[i];
36 ans += sum[pox];
37 }
38 cout<<ans<<endl;
39 }
40 return 0;
41 }
原文地址：https://www.cnblogs.com/zllwxm123/p/9573185.html