南京网络赛题解 2018

Posted 2021-01-03 weixq351

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 = 1cdot 6=6 cdot 1=2cdot 3=3cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=ab and n=ban=ban=ba are considered different if a?=ba ot = 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(Tle 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 2cdot 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

ACM-ICPC 2018 南京赛区网络预赛

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
", sum[n]);
63     }
64     return 0;
65 }