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
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4 #define mod 1000000007
5 #define inf 999999999
6 #define pi 4*atan(1)
7 //#pragma comment(linker, "/STACK:102400000,102400000")
8 int p[10]={2,3,5,7,11,13};
9 int num[20010],ji,ans;
10 struct is
11 {
12 int x;
13 int step;
14 bool operator <(const is a)const
15 {
16 return x>a.x;
17 }
18 };
19 priority_queue<is>q;
20 map<int,int>m;
21 int gcd(int x,int y)
22 {
23 return y==0?x:gcd(y,x%y);
24 }
25 void init()
26 {
27 ji=0;
28 is a;
29 a.x=1;
30 m[1]=1;
31 a.step=0;
32 q.push(a);
33 while(!q.empty())
34 {
35 is b=q.top();
36 if(b.x>1e9)
37 break;
38 q.pop();
39 if(b.step%2==0)
40 num[ji++]=b.x;
41 for(int i=0;i<6;i++)
42 {
43 is c;
44 ll gg=(ll)b.x*p[i];
45 if(gg>1e9)break;
46 c.step=b.step+1;
47 c.x=(int)gg;
48 if(c.x<=1e9&&m[c.x]==0)
49 q.push(c),m[c.x]=1;
50 }
51 }
52 }
53 void dfs(int lcm,int pos,int step,int x)
54 {
55 if(lcm>x)
56 return;
57 if(pos==6)
58 {
59 if(step==0)
60 return;
61 if(step&1)
62 ans+=x/lcm;
63 else
64 ans-=x/lcm;
65 return;
66 }
67 dfs(lcm,pos+1,step,x);
68 dfs(lcm/gcd(p[pos],lcm)*p[pos],pos+1,step+1,x);
69 }
70 int main()
71 {
72 int x,y,z,i,t;
73 init();
74 int T;
75 scanf("%d",&T);
76 while(T--)
77 {
78 scanf("%d",&x);
79 int Ans=0;
80 for(i=0;i<ji&&num[i]<=x;i++)
81 {
82 ans=0;
83 dfs(1,0,0,x/num[i]);
84 Ans+=(x/num[i]-ans);
85 }
86 printf("%d\n",Ans);
87 }
88 return 0;
89 }