B. Nastya Studies Informatics（数学）

最新推荐文章于 2020-04-24 18:36:53 发布

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该问题要求找到在给定范围内（l到r）的整数对（a, b），使得它们的最大公约数GCD(a, b)等于x，且最小公倍数LCM(a, b)等于y。通过解析题目条件，可以得出解题思路：由于x*y = a*b，可以枚举a的值，然后计算对应的b值，确保它们在给定范围内。最终计算满足条件的整数对数量。" 107446756,9058892,高可用架构：负载均衡策略与实现,"['分布式', 'nginx', '数据库', 'java', 'linux']

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Today on Informatics class Nastya learned about GCD and LCM (see links below). Nastya is very intelligent, so she solved all the tasks momentarily and now suggests you to solve one of them as well.

We define a pair of integers $(a, b)$ good, if $G C D (a, b) = x$ and $L C M (a, b) = y$ , where $G C D (a, b)$ denotes the greatest common divisor of $a$ and $b$ , and $L C M (a, b)$ denotes the least common multiple of $a$ and $b$ .

You are given two integers $x$ and $y$ . You are to find the number of good pairs of integers $(a, b)$ such that $l \leq a, b \leq r$ . Note that pairs $(a, b)$ and $(b, a)$ are considered different if $a \neq b$ .

Input

The only line contains four integers $l, r, x, y$ ( $1 \leq l \leq r \leq 10^{9}$ , $1 \leq x \leq y \leq 10^{9}$ ).

Output

In the only line print the only integer — the answer for the problem.

       Examples
      
         input
         
          Copy
         
1 2 1 2

         output
         
          Copy
         
2

         input
         
          Copy
         
1 12 1 12

         output
         
          Copy
         
4

         input
         
          Copy
         
50 100 3 30

         output
         
          Copy
         
0

Note

In the first example there are two suitable good pairs of integers $(a, b)$ : $(1, 2)$ and $(2, 1)$ .

In the second example there are four suitable good pairs of integers $(a, b)$ : $(1, 12)$ , $(12, 1)$ , $(3, 4)$ and $(4, 3)$ .

In the third example there are good pairs of integers, for example, $(3, 30)$ , but none of them fits the condition $l \leq a, b \leq r$ .

题意：a,b为两个正整数，l<= a,b <=r。x=gcd（a，b），y=lcm（a，b）。gcd是a，b的最大公约数，lcm是a，b的最小公倍数。给出l，r，x，y，求有多少种可能的a，b。

思路：如果直接暴力，利用x*y=a*b 来做的话，肯定会T！现在，我们设 a=n*m,b=m*z.那么，x=m，y=n*m*z，a*b=n*m*m*z。如果我们来枚举a的所有可能的话，1e9，太大，所以我们考虑枚举n的所有可能！！我们先来确定n的范围，最小为1，n最大值为sqrt（n*z），确定了范围后，就可以开始写了。

   #include "iostream"
    using namespace std;
    int gcd(int a,int b)
    {
        if(b==0) return a;
        return gcd(b,a%b);
    }
    int main()
    {
        int l,r,x,y,i,num,ans=0;
        cin>>l>>r>>x>>y;
        num=y/x; if(y%x!=0){cout<<"0"<<endl;return 0;}//num=n*z
        for(i=1;i*i<=num;i++){
            int j=num/i;
            if(num%i==0&&gcd(i,j)==1&&l<=i*x&&i*x<=r&&l<=j*x&&j*x<=r) ans+=i==j?1:2;
        }
        cout<<ans<<endl;
    }