Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 640    Accepted Submission(s): 207

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.

 

 

Source

 

 

Recommend

liuyiding

 

 

 

这题就是找规律。这样的题目比赛时没有快速做出来。。。只能对自己无语了。。。

 

很容易发现 所谓的 real number  就是大于4而且不是偶数的平方的偶数，或者是奇数的平方的奇数。

 

所以可以写个函数求出小于等于n的real number的个数。

 

n小于等于4，real number的个数一定为0.

(n-4)/2表示大于4的偶数的个数。但这其中包括了偶数的平方，少了奇数的平方。

所以如果 k*k<=n<(k+1)*(k+1).假如k是偶数，那么就是 (n-4)/2 ,因为偶数的平方和奇数的平方个数相等。

假如k是奇数，那么就是  (n-4)/2+1了，因为奇数的平方多一个。

 

其余就简单了。注意数据范围，用 long long

 

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;//大于4，而且不是偶数的平方数的偶数是real number//奇数的平方的奇数是real numberlong long calc(long long n)//计算小于等于n的real number的个数{    if(n<=4)return 0;    long long t=sqrt(n);    long long ans=(n-4)/2;//大于4的偶数的个数    if(t%2==0)return ans;    else return ans+1;}int main(){    int T;    long long A,B;    scanf("%d",&T);    while(T--)    {        scanf("%I64d%I64d",&A,&B);        printf("%I64d\n",calc(B)-calc(A-1));    }    return 0;}