【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[];
 int ant;
 void ResolvePrime(int n)        //分解质因数
 {
     int endd = sqrt(n);
     for (int i =  ; i <= endd ; i++)
     {
         if (n % i == )
         {
             p[ant++] = i;
             while (n % i == )
                 n /= i;
         }
     }
     if (n > )
         p[ant++] = n;
 }

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

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