求得n的因数后,简单容斥

Co-prime

    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

21 10 23 15 5

    Sample Output

Case #1: 5Case #2: 10<em>Hint</em>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        :CKbossCreated 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 #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++) cas="1;" cin="" int="" n="">>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++<<": :="" pre="" return=""></cas++<<":></sz;i++)></ll></map></set></queue></vector></cstdlib></cmath></string></cstring></cstdio></iostream>