[Number Theory][Principle of Inclusion and Exclusion]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 <= 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}.

 

Idea: The number of numbers that are relatively prime to n in the interval [a,b] can be transformed into finding the number of numbers that are relatively prime to n in [1,b] and the number that is relatively prime to n in [1,a] The number of numbers can be subtracted at the end.

The number of numbers that are coprime to n in [1,b] can be transformed into the number of numbers that are not coprime to n in [1,b], the total number minus it is the number of coprime .

To ask for the number of numbers in [1,b] that are not coprime to n, we must first know what kind of number is not coprime to n: it must be a multiple of a prime factor of n.

So first decompose n into prime factors to get several prime factors of n {fac[1], fac[2],..}, and then find out how many multiples of fac[1] in [1,b], how many The multiples of fac[2], ..., and then according to the principle of inclusion and exclusion, we can get how many numbers in [1,b] are multiples of a number in {fac[1],fac[2],..}.

The principle of inclusion and exclusion: Assuming that n has three prime factors {fac[1], fac[2], fac[3]}, then [1,b] is {fac[1], fac[2], fac[3] The number of multiples of a number in } is f(b)=n/fac[1]+n/fac[2]+n/fac[3]-n/(fac[1]*fac[2])-n /(fac[1]*fac[3])-n/(fac[2]*fac[3])+n/(fac[1]*fac[2]*fac[3]) (odd and even reduce)

 

AC code:

#include <iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;

ll factor[1000010];
ll que[1000010];
ll cnt;

void get_factor(ll n){ // Decompose prime factors (simulate short division) 
  memset(factor, 0 , sizeof (factor));
  cnt=0;
  for(ll i=2;i*i<=n;i++){
    if(n%i==0){
        factor[++cnt]=i;
        while(n%i==0) n/=i;
    }
  }
  if(n>1) factor[++cnt]=n;
}

ll fun(ll n){ // Use an array to calculate the formula 
  memset(que, 0 , sizeof (que));
  ll k=0;
  for(ll i=1;i<=cnt;i++){
    that[ ++k]= factor[i];
    ll tmp=k;
    for(int j=1;j<=tmp-1;j++) que[++k]=que[tmp]*que[j]*(-1);
  }
  ll ret=0;
  for(ll i=1;i<=k;i++) ret+=n/que[i];
  return ret;
}

intmain ()
{
    ll t;
    scanf("%lld",&t);
    ll a,b,n;
    for(ll i=1;i<=t;i++){
        scanf("%lld%lld%lld",&a,&b,&n);
        get_factor(n);
        printf("Case #%lld: %lld\n",i,b-fun(b)-(a-1-fun(a-1)));
    }
    return 0;
}