Hdu4135 -- Co-prime (Euler's Function + refresh principle)

Hdu4135 -- Co-prime (Euler's Function + refresh principle)

Co-prime Time Limit:1000 MS Memory Limit:32768KB 64bit IO Format:% I64d & % I64uSubmit StatusAppoint description: System Crawler)

Description

Given a number N, you are asked to count the number of integers between A and B random sive 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 random sive which are relatively prime to N. Follow the output format below.

Sample Input

 21 10 23 15 5 

Sample Output

 Case #1: 5Case #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}.          

Calculate the number of mutual quality between a and B.

Calculate the number of interconnectivity with n in 1 to A-1 and in 1 to B, and calculate the number of interconnectivity with n in 1 to m, respectively, first, calculate the number of numbers between 1 to m and n that are not in mutual quality.

Calculate the number of prime numbers after n decomposition, and use the binary number to represent the selection and deselection of the I-th prime number. then obtain the number contained in m. If an odd number of prime numbers is selected, the statistical results are accumulated, if it is an even number, subtract.

#include 
 
  #include 
  
   #include using namespace std ;#define LL __int64int 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<