Xtu 1242 Yada Number tolerance principle

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: Q 1[,n] interval, how many numbers, its 2,3,5,7,11,13 of the number of these factors are even

Idea: Preprocess all x to meet the 2,3,5,7,11,3 of X only, and the number is even. The number of x 13000+;

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. Sum all f (n/x)

I use the priority queue and map to process X; ll timeout; there's a place that explodes int and handles the next

1#include <bits/stdc++.h>2 using namespacestd;3 #definell Long Long4 #defineMoD 10000000075 #defineINF 9999999996 #definePi 4*atan (1)7 //#pragma COMMENT (linker, "/stack:102400000,102400000")8 intp[Ten]={2,3,5,7, One, -};9 intnum[20010],ji,ans;Ten struct  is One { A     intx; -     intStep; -     BOOL operator< (Const  isAConst the     { -         returnX>a.x; -     } - }; +priority_queue< is>Q; -map<int,int>m; + intgcdintXinty) A { at     returny==0? X:GCD (y,x%y); - } - voidInit () - { -li{0; -      isA; ina.x=1; -m[1]=1; toa.step=0; + Q.push (a); -      while(!q.empty ()) the     { *          isb=q.top (); $         if(b.x>1e9)Panax Notoginseng          Break; - Q.pop (); the         if(b.step%2==0) +num[ji++]=b.x; A          for(intI=0;i<6; i++) the         { +              isC; -ll gg= (LL) b.x*P[i]; $             if(GG&GT;1E9) Break; $c.step=b.step+1; -C.x= (int) GG; -             if(c.x<=1e9&&m[c.x]==0) theQ.push (c), m[c.x]=1; -         }Wuyi     } the } - voidDfsintLcmintPosintStepintx) Wu { -     if(lcm>x) About     return; $     if(pos==6) -     { -         if(step==0) -         return; A         if(step&1) +ans+=x/LCM; the         Else -ans-=x/LCM; $         return; the     } theDFS (lcm,pos+1, step,x); theDFS (LCM/GCD (P[POS],LCM) *p[pos],pos+1, step+1, x); the } - intMain () in { the     intx,y,z,i,t; the init (); About     intT; thescanf"%d",&T); the      while(t--) the     { +scanf"%d",&x); -         intans=0; the          for(i=0; i<ji&&num[i]<=x;i++)Bayi         { theans=0; theDfs1,0,0, x/num[i]); -ans+= (x/num[i]-ans); -         } theprintf"%d\n", Ans); the     } the     return 0; the}

Xtu 1242 Yada Number tolerance principle