hdu4279 数论（找规律）

Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1401    Accepted Submission(s): 423

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.

打出数据，找规律。东北赛的时候也是一道数论，找规律，就是简单的输入两个数字m和n，然后输出乘积~~

#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
int f[20]={0,0,0,0,0,0,1,1,2,3,4,4,5};
__int64 get(__int64 x)
{
    if(x<=12) return f[x];
    __int64 tmp=x/2;
    __int64 now=(__int64)sqrt(1.0*x);
    if(now&1) tmp-=1;
    else tmp-=2;
    return tmp;
}
int main()
{
    __int64 a,b,t;
    scanf("%I64d",&t);
    while(t--)
    {
        scanf("%I64d%I64d",&a,&b);
        printf("%I64d\n",get(b)-get(a-1));
    }
}