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 ",Ans); 87 } 88 return 0; 89 }