hdoj 4135 Co-prime 【容斥原理】

最新推荐文章于 2019-02-08 20:01:10 发布

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本文介绍了一种利用容斥原理解决质数对计数问题的算法，通过分解指定整数的所有质因数，并采用队列实现及位运算两种方式来计算在给定区间内与该数互质的整数个数。

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Co-prime

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1882 Accepted Submission(s): 722

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

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}.

容斥原理：队列实现

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<stack>
#include<algorithm>
#include<iostream>
using namespace std;
long long p[100],k;
void getp(long long n)
{
	long long i;
	k=0;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		p[k++]=i;
		while(n%i==0)
		n/=i;
	}
	if(n>1)
	p[k++]=n;
}
long long nop(long long m)
{
	long long que[100000],i,j,sum,top=0,t;
	que[top++]=-1;
	for(i=0;i<k;i++)
	{
		t=top;
		for(j=0;j<t;j++)
		{
			que[top++]=que[j]*p[i]*(-1);
		}
	}
	for(i=1,sum=0;i<top;i++)
	sum+=m/que[i];
	return sum;
}
int main()
{
	long long a,b,n;
	int t,j=1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld%lld",&a,&b,&n);
		getp(n);
		printf("Case #%d: %lld\n",j++,b-nop(b)-(a-1-nop(a-1)));
	}
	return 0;
}

更新2015.6.14

位运算：

#include <cstdio>
#include <cstring>
#define LL long long
LL p[100], k;
void getp(LL n)
{
	LL i;
	k = 0;
	for(i = 2; i*i <= n; i++)
	{
		if(n % i == 0)
		{
			p[k++] = i;
			while(n % i == 0)
			n /= i;
		}
	} 
	if(n > 1) p[k++] = n;
}
LL nop(LL m)
{
	LL i, j;
	LL use, res;
	LL sum = 0;
	for(i = 1; i < 1<<k; i++)
	{
		use = 0;
		res = 1;
		for(j = 0; j < k; j++)
		{
			if(i & (1<<j))
			{
				use++;
				res *= p[j];
			}
		}
		if(use & 1) sum += m / res;
		else sum -= m / res;
	} 
	return sum;
}
int main()
{
	int t;
	int j = 1;
	LL a, b, n;
	scanf("%d", &t);
	while(t--)
	{
		scanf("%lld%lld%lld", &a, &b, &n);
		getp(n);
		printf("Case #%d: %lld\n", j++, b-nop(b)-(a-1-nop(a-1)));
	}
	return 0;
}