Nastya Studies Informatics

Nastya Studies Informatics

 

time limit per test

1 second

 

memory limit per test

256 megabytes

 

input

standard input

output

standard output

 

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 $(a,b)good,\; if GCD(a,b)=x and LCM(a,b)=y,\; where GCD(a,b)denotes\; the greatest\; common\; divisor of a and b, and LCM(a,b)denotes\; the least\; common\; multiple of a and b.$

You are given two integers $\mathrm{x and y.\; You\; are\; to\; find\; the\; number\; of good pairs\; of\; integers (a,b) such\; that l\le a,b\le r.\; Note\; that\; pairs (a,b) and (b,a) are\; considered\; different\; if a\ne b.}$

 

Input

The only line contains four integers $\mathrm{l,r,x,y (1\le l\le r\le 109, 1\le x\le y\le 109).}$

Output

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

 

Examples

input

1 2 1 2

output

2

input

1 12 1 12

output

4

input

50 100 3 30

output

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, $(3,30),\; but\; none\; of\; them\; fits\; the\; condition l\le a,b\le rl\le a,b\le r.$

---

题意就是给你一个范围（l,r）,让你找有多少个数对（a,b）满足a和b的最大公约数是x，最小公倍数是y

思路就是对x，y进行质因数分解，这样对于每一个质因数，就会有一个上界和一个下界，而a，b对应的质因数取值也只能取上界或者下界

这样就可以把所有满足条件的数对都找出来，然后再一个个的判断是否在（l,r）范围里。

很显然数对的个数不会超过2的二十次方左右，所以不会超时。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define N 1000005
using namespace std;

long long prime[N+110],len=0,pd[N+100]= {0};

void shushu()
{
    pd[0]=pd[1]=1;

    for(int i=2; i<=N; i++)
    {
        if(pd[i]==0)
        {
            prime[len++]=i;
        }

        for(int j=0; j)
        {
            if(i*prime[j]>=N)
                break;
            pd[i*prime[j]]=1;
            if(i%prime[j]==0)
                break;
        }
    }
}

long long l,r,x,y,team[100000],team1[100000]= {0},team2[100000]= {0},c1=0,ans=0;

long long ff(long long a,long long b);
void f(long long x,long long a,long long b)
{

    if(x==c1)
    {
        if(a>=l&&a<=r&&b>=l&&b<=r)
        {
            ans++;

        }

        return;
    }
    f(x+1,a,b);


    if(team1[x]!=team2[x])
        f(x+1,a/ff(team[x],team1[x])*ff(team[x],team2[x]),b/ff(team[x],team2[x])*ff(team[x],team1[x]));


}

long long ff(long long a,long long b)
{
    long long ans=1;

    while(b>0)
    {
        if(b%2==1)
        {
            ans*=a;

        }

        a*=a;
        b/=2;

    }
    return ans;
}

int main()
{
    shushu();

    scanf("%I64d %I64d %I64d %I64d",&l,&r,&x,&y);

    if(y%x!=0)
    {
        printf("0");
        return 0;
    }

    for(int i=0; i)
    {
        if(y%prime[i]==0)
            team[c1++]=prime[i];

        while(y%prime[i]==0)
        {
            team2[c1-1]++;
            y/=prime[i];

        }

    }



    if(y>1)
    {
        team[c1++]=y;
        team2[c1-1]++;
    }


    for(int i=0; i)
    {
        while(x%team[i]==0)
        {
            team1[i]++;
            x/=team[i];
        }
    }

    long long a=1,b=1;

    for(int i=0; i)
        a*=ff(team[i],team1[i]);
    for(int i=0; i)
        b*=ff(team[i],team2[i]);

    f(0,a,b);
    printf("%I64d",ans);
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/tian-luo/p/9198206.html