Co-prime（容斥原理）

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6425    Accepted Submission(s): 2569

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

 

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lcy

AC代码：

#include<iostream>
#include<cstdio>
#include<vector>
typedef long long ll;

using namespace std;

long long solve(long long n, long long r)
{
    vector<int> p;
    p.clear();
    for(ll i = 2; i * i <= n; i++)
    {
        if(n % i == 0)
        {
            p.push_back(i);
            while(n % i == 0)
            {
                n /= i;
            }
        }
    }
    if(n > 1)
        p.push_back(n);
    ll sum = 0;
    for(ll msk = 1; msk < ((ll)1 << p.size()); msk++)
    {
        ll mult = 1, bits = 0;
        for(ll i = 0; i < p.size(); i++)
        {
            if(msk & ((ll)1 << i))
            {
                bits++;
                mult *= p[i];
            }
        }
        ll cur = r / mult;
        if(bits % 2 == 1)
            sum += cur;
        else
            sum -= cur;
    }
    return r-sum;
}

int main()
{
    int t;
    long long a, b, n;
    scanf("%d", &t);
    int cas = 1;
    while(t--)
    {
        scanf("%lld %lld %lld", &a, &b, &n);
        long long f1 = solve(n, b);
        long long f2 = solve(n, a-1);
        printf("Case #%d: %lld
", cas++, f1 - f2);
    }
    return 0;
}