【转】Hdu--4135 Co-prime

Problem Description

Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

 

Input

The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10¹⁵) and (1 <=N <= 10⁹).

 

Output

For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.

 

Sample Input

2

1 10 2

3 15 5

 

Sample Output

Case #1: 5

Case #2: 10

Hint

In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

 

Source

The Third Lebanese Collegiate Programming Contest 

代码：

/*
[a,b] 与n互质的数的个数

1~b : a1
1~a-1 : a2 

ans = a1 - a2

----> [1,k] 与 n 互质的数的个数

----> num - 与n不互质的数的个数

*/
#include<cstdio>
#include<cmath>
int p[100000];
int ant;
void ResolvePrime(int n)        //分解质因数 
{
    int endd = sqrt(n);
    for (int i = 2 ; i <= endd ; i++)
    {
        if (n % i == 0)
        {
            p[ant++] = i;
            while (n % i == 0)
                n /= i;
        }
    }
    if (n > 1)
        p[ant++] = n;
}

__int64 solve(__int64 a,int n)        //求 1～ a 不与n互质的数的个数 
{
    __int64 ans = 0;
    for (int i = 1 ; i < ((__int64)1 << ant) ; i++)
    {
        __int64 mul = 1;
        int cnt = 0;
        for (int j = 0 ; j < ant ; j++)
        {
            if (i & ((__int64)1 << j))        //选中 
            {
                cnt++;
                mul *= p[j];
            }
        }
        if (cnt & 1)
            ans += a / mul;
        else
            ans -= a / mul;
    }
    return ans;
}

int main()
{
    int T;
    int n;
    __int64 a,b;
    int Case = 1;
    scanf ("%d",&T);
    while (T--)
    {
        ant = 0;
        scanf ("%I64d %I64d %d",&a,&b,&n);
        ResolvePrime(n);
        printf ("Case #%d: %I64d
",Case++,b-(a-1)-(solve(b,n)-solve(a-1,n)));
    }
    return 0;
}