【HDU-4135】Co-prime

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Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5468    Accepted Submission(s): 2181

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

21 10 23 15 5

 

Sample Output

Case #1: 5Case #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}. 
 

 

//      互质问题

//代码如下：

/*[a,b] 与n互质的数的个数1~b : a11~a-1 : a2 ans = a1 - a2----> [1,k] 与 n 互质的数的个数----> num - 与n不互质的数的个数*/#include<cstdio>#include<cmath>int p[100000];int ant;void ResolvePrime(int n)//分解质因数 {int endd = sqrt(n);for (int i = 2 ; i <= endd ; i++){if (n % i == 0){p[ant++] = i;while (n % i == 0)n /= i;}}if (n > 1)p[ant++] = n;}__int64 solve(__int64 a,int n)//求 1～ a 不与n互质的数的个数 {__int64 ans = 0;for (int i = 1 ; i < ((__int64)1 << ant) ; i++){__int64 mul = 1;int cnt = 0;for (int j = 0 ; j < ant ; j++){if (i & ((__int64)1 << j))//选中 {cnt++;mul *= p[j];}}if (cnt & 1)ans += a / mul;elseans -= a / mul;}return ans;}int main(){int T;int n;__int64 a,b;int Case = 1;scanf ("%d",&T);while (T--){ant = 0;scanf ("%I64d %I64d %d",&a,&b,&n);ResolvePrime(n);printf ("Case #%d: %I64d\n",Case++,b-(a-1)-(solve(b,n)-solve(a-1,n)));}return 0;}