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.
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
Recommend
lcy
AC代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<vector> 4 typedef long long ll; 5 6 using namespace std; 7 8 long long solve(long long n, long long r) 9 { 10 vector<int> p; 11 p.clear(); 12 for(ll i = 2; i * i <= n; i++) 13 { 14 if(n % i == 0) 15 { 16 p.push_back(i); 17 while(n % i == 0) 18 { 19 n /= i; 20 } 21 } 22 } 23 if(n > 1) 24 p.push_back(n); 25 ll sum = 0; 26 for(ll msk = 1; msk < ((ll)1 << p.size()); msk++) 27 { 28 ll mult = 1, bits = 0; 29 for(ll i = 0; i < p.size(); i++) 30 { 31 if(msk & ((ll)1 << i)) 32 { 33 bits++; 34 mult *= p[i]; 35 } 36 } 37 ll cur = r / mult; 38 if(bits % 2 == 1) 39 sum += cur; 40 else 41 sum -= cur; 42 } 43 return r-sum; 44 } 45 46 int main() 47 { 48 int t; 49 long long a, b, n; 50 scanf("%d", &t); 51 int cas = 1; 52 while(t--) 53 { 54 scanf("%lld %lld %lld", &a, &b, &n); 55 long long f1 = solve(n, b); 56 long long f2 = solve(n, a-1); 57 printf("Case #%d: %lld\n", cas++, f1 - f2); 58 } 59 return 0; 60 }
