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HDOJ 4135 Co-prime 容斥原理

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HDOJ 4135 Co-prime 容斥原理

求得n的因數後,簡單容斥

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 <= 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

HintIn 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時52分24秒
File Name     :HDOJ4135.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include

using namespace std;

typedef long long int LL;

LL A,B,N;
vector 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>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 #"<