hdu4135--Co-prime（欧拉函数+容斥原理）

Co-prime

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Appoint description:  System Crawler  (2015-01-07)

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 <= 10  ¹⁵) and (1 <=N <= 10  ⁹).

 

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}. 
         

题目大意： 计算a到b内，与n互质的个数

分别统计1到a-1中，和1到b中与n互质的数，在相减，求1到m中与n互质的数，先求1到m中与n不互质的数的个数。

求出n分解后的质数个数，用二进制数表示第i个质数的选和不选，得到m内包含的个数，当选了奇数个质数是，统计的结果累加，为偶数个时，减去。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define LL __int64
int prim[1000000] , vis[1000000] , cnt ;
void sieve()
{
   memset(vis,0,sizeof(vis)) ;
   cnt = 0 ;
   LL i , j ;
   for(i = 2 ; i <= 1000000 ; i++)
   {
       if( !vis[i] )
       {
           prim[cnt++] = i ;
           for(j = i*i ; j < 1000000 ; j += i)
                vis[j] = 1 ;
       }
   }
}
LL p[1000] , p_num ;
LL f(LL n,LL m)
{
    LL k = n , temp , ans = 0 ;
    int i , j , num ;
    for(i = 0 , p_num = 0 ; i < cnt ; i++)
    {
        if( k % prim[i] == 0 )
        {
            p[p_num++] = prim[i] ;
        }
        while( k % prim[i] == 0 )
        {
            k /= prim[i] ;
        }
        if(k == 1)
            break ;
    }
    if( k > 1 )
        p[p_num++] = k ;
    for(i = 1 ; i < (1<<p_num) ; i++)
    {
        for(j = 0 , num = 0 , temp = 1 ; j < p_num ; j++)
        {
            if( (1<<j) & i )
            {
                temp *= p[j] ;
                num++ ;
            }
        }
        if( num % 2 )
            ans += m/temp ;
        else
            ans -= m/temp ;
    }
    return ans ;
}
int main()
{
    LL t , tt , a , b , n ;
    sieve() ;
    scanf("%I64d", &t) ;
    for(tt = 1 ; tt <= t ; tt++)
    {
        scanf("%I64d %I64d %I64d", &a, &b, &n) ;
        printf("Case #%I64d: %I64d\n", tt, (b-f(n,b))-(a-1-f(n,a-1)) );
    }
    return 0;
}