hdu 4135 二进制搞掂容斥定理 @质因子

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 193    Accepted Submission(s): 85

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 <= 10 ¹⁵) and (1 <=N <= 10 ⁹).

 

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}.

 

#include
#define maxn 90
#define LL long long
using namespace std;
LL prime[maxn]; //所存值的质因子
LL make_ans(LL num,int m){ //num个数中与m个质因子不互质的所有个数
	int ans = 0,flag,tmp,j,i;
	for(i = 1;i<(LL)(1<质因子
		}
		if(n>1){
			prime[m++] = n;
		}
		printf("Case #%d: %I64d\n",++t,(b-make_ans(b,m))- (a-1-make_ans(a-1,m))); 
	}
	return 0;
}