HDOJ 4135 Co-prime 容斥原理

求得n的因数后,简单容斥

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1798    Accepted Submission(s): 685

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}. 
    
    
    
    
     
   
   
   
   

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年03月27日 星期五 17时52分24秒
File Name     :HDOJ4135.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

LL A,B,N;
vector<LL> pr;

void getPr()
{
	LL TN=N,now=2;
	pr.clear();
	
	while(TN!=1)
	{
		if(TN%now==0)
		{
			pr.push_back(now);
			while(TN%now==0) TN/=now;
		}
		now++;
		if(now*now>TN) break;
	}
	if(TN!=1) pr.push_back(TN);
}

LL RET,RET1,RET2;

void dfs(int x,LL n,int mark)
{
	for(int i=x,sz=pr.size();i<sz;i++)
	{
		RET+=mark*(n/pr[i]);
		dfs(i+1,n/pr[i],-mark);
	}
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T,cas=1;
	cin>>T_T;
	while(T_T--)
	{
		cin>>A>>B>>N;
		getPr();
		A--; RET=A; dfs(0,A,-1); RET1=RET;
		RET=B; dfs(0,B,-1); RET2=RET;
		cout<<"Case #"<<cas++<<": "<<RET2-RET1<<endl;
	}
    
    return 0;
}