HDU 4135：Co-prime 容斥原理求(1,m)中与n互质的数的个数

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 121    Accepted Submission(s): 56

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}.

 

Source

The Third Lebanese Collegiate Programming Contest

 

Recommend

lcy

//开始系统的学习容斥原理！通常我们求１～ｎ中与ｎ互质的数的个数都是用欧拉函数！　但如果ｎ比较大或者是求１～ｍ中与ｎ互质的数的个数等等问题，要想时间效率高的话还是用容斥原理！

//本题是求[a,b]中与n互质的数的个数，可以转换成求[1,b]中与n互质的数个数减去[1,a-1]与n互质的数的个数。

#include
#include
#include
using namespace std;
#define LL long long 
#define maxn 70

LL prime[maxn];
LL make_ans(LL num,int m)
{
	LL ans=0,tmp,i,j,flag;
	for(i=1;i<(LL)(1<1)
			prime[m++]=n;
		printf("Case #%d: %I64d\n",++t,(b-make_ans(b,m))-(a-1-make_ans(a-1,m)));
	}
	return 0;
}