HDU 4135 Co-prime 解题报告（因式分解 + 容斥原理）

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1111    Accepted Submission(s): 405

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

 

解题报告：简单来说就是求区间[a, b]中于n互为素数的数的个数。我们知道如果求区间[1, n]与n互为素数的数的个数，可以使用欧拉定理。但是本题中无法使用。

不过可以简单转化一下，题目就是求[1, b]中互素的个数减去[1, a-1]中互素的个数。此时我们可以使用容斥原理。

即[1, b]中与2互为约数的数共有b/2个，与3互为约数的数的个数有b/3个。同时与2，3互为约数的数的个数就是b/2+b/3-b/lcm(2,3)。

把n因式分解分解。简单计算可以知道，前10个素数相乘大于6*10^10，n的范围是10^9，也就是说n的素数因子不超过10个。那么复杂度就不会超过2^10*10。完全可以接受。

代码如下：

#include 
#include 
#include 
using namespace std;

typedef long long LL;

int prime[30];
int top;
int cas;

LL co_prime(LL val)
{
	LL ans = 0;

	for (int i = 1; i < (1 << top); i++)
	{
		LL tmp = 1, flag = 0;

		for (int j = 0; j < top; j++) if (i&(1 << j))
			tmp *= prime[j], flag++;

		if (flag & 1)
			ans += val / tmp;
		else
			ans -= val / tmp;
	}

	return val - ans;
}

void work()
{
	LL a, b;
	int n;
	scanf("%I64d%I64d%d", &a, &b, &n);

	top = 0;
	for (int i = 2; i*i <= n; i++) if (n%i==0)
	{
		prime[top++] = i;
		while (n%i == 0) n /= i;
	}

	if (n > 1)
		prime[top++] = n;

	printf("Case #%d: %I64d\n", ++cas, co_prime(b) - co_prime(a-1));
}

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
		work();
}