hdu 4279 Number

http://acm.hdu.edu.cn/showproblem.php?pid=4279

Problem Description
  Here are two numbers A and B (0 < A <= B). If B cannot be divisible by A, and A and B are not co-prime numbers, we define A as a special number of B.
  For each x, f(x) equals to the amount of x’s special numbers.
  For example, f(6)=1, because 6 only have one special number which is 4. And f(12)=3, its special numbers are 8,9,10.
  When f(x) is odd, we consider x as a real number.
  Now given 2 integers x and y, your job is to calculate how many real numbers are between them.
 
Input
  In the first line there is an integer T (T <= 2000), indicates the number of test cases. Then T line follows, each line contains two integers x and y (1 <= x <= y <= 2^63-1) separated by a single space.
 
Output
  Output the total number of real numbers.
 
Sample Input
2 1 1 1 10
 
Sample Output
0 4
Hint
For the second case, the real numbers are 6,8,9,10.
 
 
这题实在是规律难找,参考了几位大神的代码
 1 #include <cstdio>
 2  #include <cmath>
 3  #include <cstring>
 4  #include <algorithm>
 5  #include <iostream>
 6  
 7  using namespace std;
 8  
 9  long long a, b, t;
10  long long s1, s2;
11  
12  int main()
13  {
14       int T;
15       scanf("%d", &T);
16       while(T--) {
17            cin>>a>>b;
18            --a;
19            s1 = a >> 1;
20            s2 = b >> 1;
21  
22            t = sqrt(a);
23            s1 -= t >> 1;
24            s1 += t-(t>>1);
25  
26            t = sqrt(b);
27            s2 -= t >> 1;
28            s2 += t-(t>>1);
29  
30            if (a >= 1) s1 -= 1;
31            if (a >= 2) s1 -= 1;
32            if (b >= 1) s2 -= 1;
33            if (b >= 2) s2 -= 1;
34            s2 -= s1;
35            cout<<s2<<endl;
36       }
37       return 0;
38  }

代码(2):参考自http://blog.csdn.net/kk303/article/details/7960986

 1 #include<iostream>
 2 #include<cmath>
 3 using namespace std;      
 4 int t,n,s[20]={0,0,0,0,0,0,1,1,2,3,4,4,5};   
 5 __int64 a,b,p1,p2;
 6 __int64 getsum(__int64 x)
 7 {
 8      if (x<=12) return s[x];
 9      __int64 m,k;
10      m=x/2; 
11      k=(__int64)sqrt(x);
12      if (k%2==0) m-=2;
13         else m-=1;
14      return m;
15 }
16 int main()
17 {  
18     // freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout); 
19      scanf("%d",&t);
20      while (t--)
21      { 
22             scanf("%I64d%I64d",&a,&b);
23             a--;
24             p1=getsum(a);  
25             p2=getsum(b);
26             printf("%I64d\n",p2-p1);
27      }
28      return 0;
29 }
原文地址:https://www.cnblogs.com/crazyapple/p/2679469.html