J. Sum
A square-free integer is an integer which is indivisible by any square number except 111. For example, 6=2⋅36 = 2 \cdot 36=2⋅3 is square-free, but 12=22⋅312 = 2^2 \cdot 312=22⋅3 is not, because 222^222 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⋅6=6⋅1=2⋅3=3⋅2,n=ab6 = 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=ban=ba are considered different if a?=ba \not = ba?=b. f(n)f(n)f(n) is the number of decomposition ways that n=abn=abn=ab such that aaa and bbb are square-free integers. The problem is calculating ∑i=1nf(i)\sum_{i = 1}^nf(i)∑i=1n?f(i).
Input
The first line contains an integer T(T≤20)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≤2⋅107)n(n \le 2\cdot 10^7)n(n≤2⋅107).
Output
For each test case, print the answer ∑i=1nf(i)\sum_{i = 1}^n f(i)∑i=1n?f(i).
Hint
∑i=18f(i)=f(1)+?+f(8)\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=1+2+2+1+2+4+2+0=14.
样例输入
2 5 8
样例输出
8 14
题目来源
解析：题目求和1-n的各个数的非平方因子的和，与质数有关，或者用积性函数线性筛。
提交验证代码处：https://nanti.jisuanke.com/t/30999
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define LL long long
4
5 int t;
6 LL n;
7 const int maxn = 22000000;
8 bool tag[maxn];
9 int p[maxn/10];
10 int cnt;
11 LL sum[maxn];
12 void GetPrime()
13 {
14 cnt = 0;
15 sum[1] = 1;
16 for(int i = 2; i < maxn; i++)
17 {
18 if(!tag[i])
19 {
20 p[cnt++] = i;
21 sum[i] = 2;
22 }
23 for(int j = 0; j < cnt && p[j] * i < maxn; j++)
24 {
25 tag[i * p[j]] = 1;
26 if(i % p[j] == 0)
27 {
28 int qq = 0, xx = i;
29 LL v = 1;
30 while (xx%p[j]==0)
31 {
32 qq ++;
33 v *= p[j];
34 xx/=p[j];
35 }
36 v *= p[j];
37 if (qq >= 2)
38 {
39 sum[i*p[j]] = 0;
40 }
41 else if(qq == 1)
42 {
43 sum[i*p[j]] = sum[i] / 2;
44 }
45 break;
46 }
47 sum[i*p[j]] = sum[i] * 2;
48 }
49 }
50 for (int i = 1; i < maxn; i++)
51 {
52 sum[i] += sum[i-1];
53 }
54 }
55 int main()
56 {
57 GetPrime();
58 cin >> t;
59 while(t--)
60 {
61 scanf("%lld", &n);
62 printf("%lld\n", sum[n]);
63 }
64 return 0;
65 }
原文地址：https://www.cnblogs.com/weixq351/p/9573604.html