Number HDU

Number HDU - 4279

　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.

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--){        long long int n,m,x1,x2;;        scanf("%lld%lld",&n,&m);        long long int tt=n-1;        if (tt==0||tt==1){            x1=0;        }        else{        long double qq=tt;            long long int ss=sqrt(qq);            if (ss%2==0){                x1=(n-1)/2-2;            }            else{                x1=(n-1)/2-1;            }        }        if (m==0||m==1){            x2=0;        }        else{        long double qq=m;            long long int sss=sqrt(qq);            if (sss%2==0){                x2=m/2-2;            }            else{                x2=m/2-1;            }        }        printf("%lld\n",x2-x1);    }    return 0;}