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
代码：
1 /* 2 [a,b] 与n互质的数的个数 3 4 1~b : a1 5 1~a-1 : a2 6 7 ans = a1 - a2 8 9 ----> [1,k] 与 n 互质的数的个数 10 11 ----> num - 与n不互质的数的个数 12 13 */ 14 #include<cstdio> 15 #include<cmath> 16 int p[100000]; 17 int ant; 18 void ResolvePrime(int n) //分解质因数 19 { 20 int endd = sqrt(n); 21 for (int i = 2 ; i <= endd ; i++) 22 { 23 if (n % i == 0) 24 { 25 p[ant++] = i; 26 while (n % i == 0) 27 n /= i; 28 } 29 } 30 if (n > 1) 31 p[ant++] = n; 32 } 33 34 __int64 solve(__int64 a,int n) //求 1～ a 不与n互质的数的个数 35 { 36 __int64 ans = 0; 37 for (int i = 1 ; i < ((__int64)1 << ant) ; i++) 38 { 39 __int64 mul = 1; 40 int cnt = 0; 41 for (int j = 0 ; j < ant ; j++) 42 { 43 if (i & ((__int64)1 << j)) //选中 44 { 45 cnt++; 46 mul *= p[j]; 47 } 48 } 49 if (cnt & 1) 50 ans += a / mul; 51 else 52 ans -= a / mul; 53 } 54 return ans; 55 } 56 57 int main() 58 { 59 int T; 60 int n; 61 __int64 a,b; 62 int Case = 1; 63 scanf ("%d",&T); 64 while (T--) 65 { 66 ant = 0; 67 scanf ("%I64d %I64d %d",&a,&b,&n); 68 ResolvePrime(n); 69 printf ("Case #%d: %I64d ",Case++,b-(a-1)-(solve(b,n)-solve(a-1,n))); 70 } 71 return 0; 72 }