hdu 4135 Co-prime (容斥定理)

Co-prime
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7679    Accepted Submission(s): 3032

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}.

C/C++：

#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

long long t, a, b, n;

vector <long long> v;

void get_prime()
{
    v.clear();
    for (long long i = 2; i * i <= n; ++ i)
    {
        if (n % i == 0) v.push_back(i);
        while (n % i == 0)
            n /= i;
    }
    if (n > 1) v.push_back(n);
}

long long num(long long m)
{
    long long sum = 0, top = 1, Q[1000] = {-1};
    for (long long i = 0; i < v.size(); ++ i)
    {
        long long temp = top;
        for (long long j = 0; j < temp; ++ j)
            Q[top ++] = -1 * v[i] * Q[j];
    }
    for (long long i = 1; i < top; ++ i)
        sum += (m / Q[i]);
    return sum;
}

int main()
{
    scanf("%lld", &t);
    for (int i = 1; i <= t; ++ i)
    {
        scanf("%lld%lld%lld", &a, &b, &n);
        get_prime();
        printf("Case #%d: %lld
", i, b - num(b) - (a - 1 - num(a - 1)));
    }
}