ACM-ICPC 2018 南京赛区网络预赛 J.Sum

Sum

A square-free integer is an integer which is indivisible by any square number except $2^2 cdot 312=22⋅3 is not, because 2^222 is a square number. Some integers could be decomposed into product of two square-free integers, there may be more than one decomposition ways. For example, 6 = 1cdot 6=6 cdot 1=2cdot 3=3cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=ab and n=ban=ba are considered different if a ot = ba̸=b. f(n)f(n) is the number of decomposition ways that n=abn=ab such that aa and bb are square-free integers. The problem is calculating sum_{i = 1}^nf(i)∑i=1nf(i).$

Input

The first line contains an integer

For each test case, there first line has a integer $10^7)n(n≤2⋅107).$

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

题目来源

ACM-ICPC 2018 南京赛区网络预赛

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 }