XTU 1242 Yada number allowance

Yada Numberproblem Description:

Every positive integer can is expressed by multiplication of prime integers. Duoxida says an integer was a yada number if the total amount of 2,3,5,7,11,13 in it prime factors is even.

For instance, 18=2 * 3 * 3 are not a yada number since the sum of amount of 2, 3 are 3, an odd number; While-2 * 5 * is a yada number since the sum of amount of 2, 5 is 2, a even number that satifies the definition O f yada number.

Now, Duoxida wonders how many yada number is among all integers in [1,n].

Input

The first line contains an integer T (no more than) which indicating the number of test cases. In the following T lines containing a integer n. ()

Output

For each case, output the answer on one single line.

Sample Input

2
18
21st

Sample Output

9
11
??????????

Test instructions

Give you a n, ask you 1 to n how many of the number of satisfied factor is the number of 2,3,5,7,11,13 is an even number

Exercises

All x is preprocessed to meet the 2,3,5,7,11,3 of the X, and the number is even. The number of x is slightly greater than 10000

Note that 0 cases are added, that is, 1.

For a number n, the enumeration of all x, for a x,f (n/x) to find out [1,n/x] does not contain the number of 2,3,5,7,11,13 as a factor, this is the classic repulsion problem.

Finally, all F (n/x) sums can be

#include <bits/stdc++.h>using namespacestd;Const intN = 3e6+ -, M = 1e6+Ten, mod = 1e9+7, INF =1e9;typedefLong Longll;Constll MAXN =1e9;intCNT =0, Ans,n;ll b[n];intA[] = {2,3,5,7, One, -};ll gcd (ll a,ll b) {returnb==0? A:GCD (b,a%b);}voidDFS (ll X,intFintnum) {    if(num==6) {        if(!f) b[cnt++] =x; return ; }     while(x<=MAXN) {DFS (X,f,num+1); X*=A[num]; F^=1; }}voidinit () {DFS (1,0,0); Sort (b,b+cnt);}voidINCLU (intIintnum,ll tmp) {    if(tmp>n)return ; if(i>=6) {        if(num==0) ans =0; Else {            if(num&1) ans = ans+n/tmp; ElseAns = ans-n/tmp; }        return ; } inclu (I+1, num,tmp); INCLU (i+1, num+1, tmp*a[i]/gcd (Tmp,a[i]));}voidsolve () {intAns =0; scanf ("%d",&N); intTM =N;  for(intI=0; i<cnt&&b[i]<=tm;i++) {n= tm/B[i]; Ans=0; INCLU (0,0,1); Ans+ = (N-ans); } printf ("%d\n", Ans);}intMain () {intT; CNT=0;    Init (); scanf ("%d",&T);  while(t--) {solve (); }    return 0;}

XTU 1242 Yada number allowance