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++：

 1 #include <map>
 2 #include <queue>
 3 #include <cmath>
 4 #include <vector>
 5 #include <string>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <climits>
 9 #include <iostream>
10 #include <algorithm>
11 #define INF 0x3f3f3f3f
12 using namespace std;
13 
14 long long t, a, b, n;
15 
16 vector <long long> v;
17 
18 void get_prime()
19 {
20     v.clear();
21     for (long long i = 2; i * i <= n; ++ i)
22     {
23         if (n % i == 0) v.push_back(i);
24         while (n % i == 0)
25             n /= i;
26     }
27     if (n > 1) v.push_back(n);
28 }
29 
30 long long num(long long m)
31 {
32     long long sum = 0, top = 1, Q[1000] = {-1};
33     for (long long i = 0; i < v.size(); ++ i)
34     {
35         long long temp = top;
36         for (long long j = 0; j < temp; ++ j)
37             Q[top ++] = -1 * v[i] * Q[j];
38     }
39     for (long long i = 1; i < top; ++ i)
40         sum += (m / Q[i]);
41     return sum;
42 }
43 
44 int main()
45 {
46     scanf("%lld", &t);
47     for (int i = 1; i <= t; ++ i)
48     {
49         scanf("%lld%lld%lld", &a, &b, &n);
50         get_prime();
51         printf("Case #%d: %lld\n", i, b - num(b) - (a - 1 - num(a - 1)));
52     }
53 }