A square-free integer is an integer which is indivisible by any square number except 1. For example, 6 = 2 * 3,6=2*3 is square-free, but 12 = 2^2 * 3,12=2^2*3 is not,
because 2^2 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=2* 3=3*2 6=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 T(T≤20), denoting the number of test cases.
For each test case, there first line has a integer 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
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1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstdlib> 5 #include <cstring> 6 #include <string> 7 #include <deque> 8 #include <set> 9 #include <queue> 10 #include <cmath> 11 #define ll long long 12 #define M 20000004 13 #define P pair<int,int> 14 using namespace std; 15 bool vis[M]; 16 int sum[M]; 17 void init() 18 { 19 int u=sqrt(M+0.5); 20 for(int i=1;i<=M;i++) vis[i]=1; 21 for(int i=2;i<=u;i++) 22 { 23 int k=i*i; 24 for(int j=k;j<M;j+=k) 25 { 26 if(vis[j]) vis[j]=0; 27 } 28 } 29 /* 30 1 2 3 4 5 6 7 8 31 1 2 3 3 4 5 6 6 32 */ 33 for(int i=1;i<M;i++) 34 { 35 if(vis[i]){ 36 sum[i]=sum[i-1]+1; 37 } 38 else sum[i]=sum[i-1]; 39 } 40 } 41 int t,n; 42 int main() 43 { 44 init(); 45 scanf("%d",&t); 46 while(t--) 47 { 48 scanf("%d",&n); 49 int u=sqrt(n+0.5); 50 ll ans=0; 51 /* 52 n==8 53 i==1 : 1*8最大是8. 54 (1,2),(1,3),(1,5),(1,6),(1,7) 55 (2,1),(3,1),(5,1),(6,1),(7,1) 56 (1,1) 57 那么 2*5+1==11 58 i==2 : 2*4最大是 4 59 (2,3) 60 (3,2) 61 (2,2) 62 1*2+1==3 63 11+3==14 64 */ 65 for(int i=1;i<=u;i++){ 66 if(!vis[i]) continue; 67 int y=n/i; 68 ans+=(ll)(sum[y]-sum[i])*2+1; 69 } 70 printf("%lld ",ans); 71 } 72 return 0; 73 }