xtu 1242 Yada Number 容斥原理

Yada Number

Problem Description:

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

For instance, 18=2 * 3 * 3 is not a yada number since the sum of amount of 2, 3 is 3, an odd number; while 170 = 2 * 5 * 17 is a yada number since the sum of amount of 2, 5 is 2, a even number that satifies the definition of yada number.

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

Input

The first line contains a integer T(no more than 50) which indicating the number of test cases. In the following T lines containing a integer n. ()

Output

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

Sample Input

2
18
21

Sample Output

9
11

题意：问1[,n]区间中，有多少个数，它的2,3,5,7,11,13的这几个因子数目之和为偶数

思路：预处理出所有的x，满足x只含有2,3,5,7,11,3这几个质因子，且数目为偶数。x的数目13000+；

对于一个数n，枚举所有的x，对于一个x，f(n/x)即求出[1,n/x]中不含有2,3,5，7,11,13作为因子的数有多少个，这个是经典的容斥问题。对所有的f(n/x)求和即可

　　　　我用优先队列和map处理x；全用ll超时；有个地方会爆int，处理了下

 #include<bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define mod 1000000007
 #define inf 999999999
 #define pi 4*atan(1)
 //#pragma comment(linker, "/STACK:102400000,102400000")
 int p[]={,,,,,};
 int num[],ji,ans;
 struct is
 {
     int x;
     int step;
     bool operator <(const is a)const
     {
         return x>a.x;
     }
 };
 priority_queue<is>q;
 map<int,int>m;
 int gcd(int x,int y)
 {
     return y==?x:gcd(y,x%y);
 }
 void init()
 {
     ji=;
     is a;
     a.x=;
     m[]=;
     a.step=;
     q.push(a);
     while(!q.empty())
     {
         is b=q.top();
         if(b.x>1e9)
         break;
         q.pop();
         if(b.step%==)
         num[ji++]=b.x;
         for(int i=;i<;i++)
         {
             is c;
             ll gg=(ll)b.x*p[i];
             if(gg>1e9)break;
             c.step=b.step+;
             c.x=(int)gg;
             if(c.x<=1e9&&m[c.x]==)
             q.push(c),m[c.x]=;
         }
     }
 }
 void dfs(int lcm,int pos,int step,int x)
 {
     if(lcm>x)
     return;
     if(pos==)
     {
         if(step==)
         return;
         if(step&)
         ans+=x/lcm;
         else
         ans-=x/lcm;
         return;
     }
     dfs(lcm,pos+,step,x);
     dfs(lcm/gcd(p[pos],lcm)*p[pos],pos+,step+,x);
 }
 int main()
 {
     int x,y,z,i,t;
     init();
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&x);
         int Ans=;
         for(i=;i<ji&&num[i]<=x;i++)
         {
             ans=;
             dfs(,,,x/num[i]);
             Ans+=(x/num[i]-ans);
         }
         printf("%d\n",Ans);
     }
     return ;
 }