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求得n的因数后,简单容斥

Co-prime

2
1 10 2
3 15 5

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;
}

HDOJ 4135 Co-prime 容斥原理

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原文地址：http://blog.csdn.net/ck_boss/article/details/44681377