Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2936 Accepted Submission(s): 805

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

2012 ACM/ICPC Asia Regional Tianjin Online

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liuyiding

题意：给出一个f(x),表示不大于x的正整数里，不整除x且跟x有大于1的公约数的数的个数。

定义F(x),为不大于x的正整数里，满足f(x)的值为奇数的数的个数。题目就是求这个F(x)。

代码例如以下：（用C++提交WA了无数次，用G++一遍过）

#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
//大于4。并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 number
__int64 calc(__int64 n)//计算小于等于n的real number的个数
{
    if(n<=4)
		return 0;
    __int64 t=sqrt(n*1.0);
    __int64 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;
    __int64 A,B;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d",&A,&B);
        printf("%I64d
",calc(B)-calc(A-1));
    }
    return 0;
}